64bit IEEE 754: Decimal -> Double Precision Floating Point Binary: -1 036.699 999 999 999 818 101 059 645 43 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number -1 036.699 999 999 999 818 101 059 645 43(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-1 036.699 999 999 999 818 101 059 645 43| = 1 036.699 999 999 999 818 101 059 645 43

2. First, convert to binary (in base 2) the integer part: 1 036.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 1 036 ÷ 2 = 518 + 0;
  • 518 ÷ 2 = 259 + 0;
  • 259 ÷ 2 = 129 + 1;
  • 129 ÷ 2 = 64 + 1;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.


1 036(10) =

100 0000 1100(2)


4. Convert to binary (base 2) the fractional part: 0.699 999 999 999 818 101 059 645 43.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.699 999 999 999 818 101 059 645 43 × 2 = 1 + 0.399 999 999 999 636 202 119 290 86;
  • 2) 0.399 999 999 999 636 202 119 290 86 × 2 = 0 + 0.799 999 999 999 272 404 238 581 72;
  • 3) 0.799 999 999 999 272 404 238 581 72 × 2 = 1 + 0.599 999 999 998 544 808 477 163 44;
  • 4) 0.599 999 999 998 544 808 477 163 44 × 2 = 1 + 0.199 999 999 997 089 616 954 326 88;
  • 5) 0.199 999 999 997 089 616 954 326 88 × 2 = 0 + 0.399 999 999 994 179 233 908 653 76;
  • 6) 0.399 999 999 994 179 233 908 653 76 × 2 = 0 + 0.799 999 999 988 358 467 817 307 52;
  • 7) 0.799 999 999 988 358 467 817 307 52 × 2 = 1 + 0.599 999 999 976 716 935 634 615 04;
  • 8) 0.599 999 999 976 716 935 634 615 04 × 2 = 1 + 0.199 999 999 953 433 871 269 230 08;
  • 9) 0.199 999 999 953 433 871 269 230 08 × 2 = 0 + 0.399 999 999 906 867 742 538 460 16;
  • 10) 0.399 999 999 906 867 742 538 460 16 × 2 = 0 + 0.799 999 999 813 735 485 076 920 32;
  • 11) 0.799 999 999 813 735 485 076 920 32 × 2 = 1 + 0.599 999 999 627 470 970 153 840 64;
  • 12) 0.599 999 999 627 470 970 153 840 64 × 2 = 1 + 0.199 999 999 254 941 940 307 681 28;
  • 13) 0.199 999 999 254 941 940 307 681 28 × 2 = 0 + 0.399 999 998 509 883 880 615 362 56;
  • 14) 0.399 999 998 509 883 880 615 362 56 × 2 = 0 + 0.799 999 997 019 767 761 230 725 12;
  • 15) 0.799 999 997 019 767 761 230 725 12 × 2 = 1 + 0.599 999 994 039 535 522 461 450 24;
  • 16) 0.599 999 994 039 535 522 461 450 24 × 2 = 1 + 0.199 999 988 079 071 044 922 900 48;
  • 17) 0.199 999 988 079 071 044 922 900 48 × 2 = 0 + 0.399 999 976 158 142 089 845 800 96;
  • 18) 0.399 999 976 158 142 089 845 800 96 × 2 = 0 + 0.799 999 952 316 284 179 691 601 92;
  • 19) 0.799 999 952 316 284 179 691 601 92 × 2 = 1 + 0.599 999 904 632 568 359 383 203 84;
  • 20) 0.599 999 904 632 568 359 383 203 84 × 2 = 1 + 0.199 999 809 265 136 718 766 407 68;
  • 21) 0.199 999 809 265 136 718 766 407 68 × 2 = 0 + 0.399 999 618 530 273 437 532 815 36;
  • 22) 0.399 999 618 530 273 437 532 815 36 × 2 = 0 + 0.799 999 237 060 546 875 065 630 72;
  • 23) 0.799 999 237 060 546 875 065 630 72 × 2 = 1 + 0.599 998 474 121 093 750 131 261 44;
  • 24) 0.599 998 474 121 093 750 131 261 44 × 2 = 1 + 0.199 996 948 242 187 500 262 522 88;
  • 25) 0.199 996 948 242 187 500 262 522 88 × 2 = 0 + 0.399 993 896 484 375 000 525 045 76;
  • 26) 0.399 993 896 484 375 000 525 045 76 × 2 = 0 + 0.799 987 792 968 750 001 050 091 52;
  • 27) 0.799 987 792 968 750 001 050 091 52 × 2 = 1 + 0.599 975 585 937 500 002 100 183 04;
  • 28) 0.599 975 585 937 500 002 100 183 04 × 2 = 1 + 0.199 951 171 875 000 004 200 366 08;
  • 29) 0.199 951 171 875 000 004 200 366 08 × 2 = 0 + 0.399 902 343 750 000 008 400 732 16;
  • 30) 0.399 902 343 750 000 008 400 732 16 × 2 = 0 + 0.799 804 687 500 000 016 801 464 32;
  • 31) 0.799 804 687 500 000 016 801 464 32 × 2 = 1 + 0.599 609 375 000 000 033 602 928 64;
  • 32) 0.599 609 375 000 000 033 602 928 64 × 2 = 1 + 0.199 218 750 000 000 067 205 857 28;
  • 33) 0.199 218 750 000 000 067 205 857 28 × 2 = 0 + 0.398 437 500 000 000 134 411 714 56;
  • 34) 0.398 437 500 000 000 134 411 714 56 × 2 = 0 + 0.796 875 000 000 000 268 823 429 12;
  • 35) 0.796 875 000 000 000 268 823 429 12 × 2 = 1 + 0.593 750 000 000 000 537 646 858 24;
  • 36) 0.593 750 000 000 000 537 646 858 24 × 2 = 1 + 0.187 500 000 000 001 075 293 716 48;
  • 37) 0.187 500 000 000 001 075 293 716 48 × 2 = 0 + 0.375 000 000 000 002 150 587 432 96;
  • 38) 0.375 000 000 000 002 150 587 432 96 × 2 = 0 + 0.750 000 000 000 004 301 174 865 92;
  • 39) 0.750 000 000 000 004 301 174 865 92 × 2 = 1 + 0.500 000 000 000 008 602 349 731 84;
  • 40) 0.500 000 000 000 008 602 349 731 84 × 2 = 1 + 0.000 000 000 000 017 204 699 463 68;
  • 41) 0.000 000 000 000 017 204 699 463 68 × 2 = 0 + 0.000 000 000 000 034 409 398 927 36;
  • 42) 0.000 000 000 000 034 409 398 927 36 × 2 = 0 + 0.000 000 000 000 068 818 797 854 72;
  • 43) 0.000 000 000 000 068 818 797 854 72 × 2 = 0 + 0.000 000 000 000 137 637 595 709 44;
  • 44) 0.000 000 000 000 137 637 595 709 44 × 2 = 0 + 0.000 000 000 000 275 275 191 418 88;
  • 45) 0.000 000 000 000 275 275 191 418 88 × 2 = 0 + 0.000 000 000 000 550 550 382 837 76;
  • 46) 0.000 000 000 000 550 550 382 837 76 × 2 = 0 + 0.000 000 000 001 101 100 765 675 52;
  • 47) 0.000 000 000 001 101 100 765 675 52 × 2 = 0 + 0.000 000 000 002 202 201 531 351 04;
  • 48) 0.000 000 000 002 202 201 531 351 04 × 2 = 0 + 0.000 000 000 004 404 403 062 702 08;
  • 49) 0.000 000 000 004 404 403 062 702 08 × 2 = 0 + 0.000 000 000 008 808 806 125 404 16;
  • 50) 0.000 000 000 008 808 806 125 404 16 × 2 = 0 + 0.000 000 000 017 617 612 250 808 32;
  • 51) 0.000 000 000 017 617 612 250 808 32 × 2 = 0 + 0.000 000 000 035 235 224 501 616 64;
  • 52) 0.000 000 000 035 235 224 501 616 64 × 2 = 0 + 0.000 000 000 070 470 449 003 233 28;
  • 53) 0.000 000 000 070 470 449 003 233 28 × 2 = 0 + 0.000 000 000 140 940 898 006 466 56;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (losing precision...)


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.699 999 999 999 818 101 059 645 43(10) =

0.1011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0000 0000 0000 0(2)


6. Positive number before normalization:

1 036.699 999 999 999 818 101 059 645 43(10) =

100 0000 1100.1011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0000 0000 0000 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 10 positions to the left, so that only one non zero digit remains to the left of it:


1 036.699 999 999 999 818 101 059 645 43(10) =

100 0000 1100.1011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0000 0000 0000 0(2) =

100 0000 1100.1011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0000 0000 0000 0(2) × 20 =

1.0000 0011 0010 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 0000 0000 000(2) × 210


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): 10

Mantissa (not normalized):
1.0000 0011 0010 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 0000 0000 000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

10 + 2(11-1) - 1 =

(10 + 1 023)(10) =

1 033(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 033 ÷ 2 = 516 + 1;
  • 516 ÷ 2 = 258 + 0;
  • 258 ÷ 2 = 129 + 0;
  • 129 ÷ 2 = 64 + 1;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1033(10) =

100 0000 1001(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 0000 0011 0010 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 000 0000 0000 =

0000 0011 0010 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
100 0000 1001

Mantissa (52 bits) =
0000 0011 0010 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100


The base ten decimal number -1 036.699 999 999 999 818 101 059 645 43 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
1 - 100 0000 1001 - 0000 0011 0010 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100

(64 bits IEEE 754)

  • Sign (1 bit):

    • 1

      63

  • Exponent (11 bits):

    • 1

      62

    • 0

      61

    • 0

      60

    • 0

      59

    • 0

      58

    • 0

      57

    • 0

      56

    • 1

      55

    • 0

      54

    • 0

      53

    • 1

      52

  • Mantissa (52 bits):

    • 0

      51

    • 0

      50

    • 0

      49

    • 0

      48

    • 0

      47

    • 0

      46

    • 1

      45

    • 1

      44

    • 0

      43

    • 0

      42

    • 1

      41

    • 0

      40

    • 1

      39

    • 1

      38

    • 0

      37

    • 0

      36

    • 1

      35

    • 1

      34

    • 0

      33

    • 0

      32

    • 1

      31

    • 1

      30

    • 0

      29

    • 0

      28

    • 1

      27

    • 1

      26

    • 0

      25

    • 0

      24

    • 1

      23

    • 1

      22

    • 0

      21

    • 0

      20

    • 1

      19

    • 1

      18

    • 0

      17

    • 0

      16

    • 1

      15

    • 1

      14

    • 0

      13

    • 0

      12

    • 1

      11

    • 1

      10

    • 0

      9

    • 0

      8

    • 1

      7

    • 1

      6

    • 0

      5

    • 0

      4

    • 1

      3

    • 1

      2

    • 0

      1

    • 0

      0

Number -1 036.699 999 999 999 818 101 059 645 44 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

Number -1 036.699 999 999 999 818 101 059 645 42 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

Convert to 64 bit double precision IEEE 754 binary floating point representation standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

The latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation

Number -1 036.699 999 999 999 818 101 059 645 43 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Nov 30 16:53 UTC (GMT)
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All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =

    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100