64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 3.141 592 653 589 793 115 997 963 468 544 185 161 590 54 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 3.141 592 653 589 793 115 997 963 468 544 185 161 590 54(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. First, convert to binary (in base 2) the integer part: 3.
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;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

2. 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.


3(10) =

11(2)


3. Convert to binary (base 2) the fractional part: 0.141 592 653 589 793 115 997 963 468 544 185 161 590 54.

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) 0.141 592 653 589 793 115 997 963 468 544 185 161 590 54 × 2 = 0 + 0.283 185 307 179 586 231 995 926 937 088 370 323 181 08;
  • 2) 0.283 185 307 179 586 231 995 926 937 088 370 323 181 08 × 2 = 0 + 0.566 370 614 359 172 463 991 853 874 176 740 646 362 16;
  • 3) 0.566 370 614 359 172 463 991 853 874 176 740 646 362 16 × 2 = 1 + 0.132 741 228 718 344 927 983 707 748 353 481 292 724 32;
  • 4) 0.132 741 228 718 344 927 983 707 748 353 481 292 724 32 × 2 = 0 + 0.265 482 457 436 689 855 967 415 496 706 962 585 448 64;
  • 5) 0.265 482 457 436 689 855 967 415 496 706 962 585 448 64 × 2 = 0 + 0.530 964 914 873 379 711 934 830 993 413 925 170 897 28;
  • 6) 0.530 964 914 873 379 711 934 830 993 413 925 170 897 28 × 2 = 1 + 0.061 929 829 746 759 423 869 661 986 827 850 341 794 56;
  • 7) 0.061 929 829 746 759 423 869 661 986 827 850 341 794 56 × 2 = 0 + 0.123 859 659 493 518 847 739 323 973 655 700 683 589 12;
  • 8) 0.123 859 659 493 518 847 739 323 973 655 700 683 589 12 × 2 = 0 + 0.247 719 318 987 037 695 478 647 947 311 401 367 178 24;
  • 9) 0.247 719 318 987 037 695 478 647 947 311 401 367 178 24 × 2 = 0 + 0.495 438 637 974 075 390 957 295 894 622 802 734 356 48;
  • 10) 0.495 438 637 974 075 390 957 295 894 622 802 734 356 48 × 2 = 0 + 0.990 877 275 948 150 781 914 591 789 245 605 468 712 96;
  • 11) 0.990 877 275 948 150 781 914 591 789 245 605 468 712 96 × 2 = 1 + 0.981 754 551 896 301 563 829 183 578 491 210 937 425 92;
  • 12) 0.981 754 551 896 301 563 829 183 578 491 210 937 425 92 × 2 = 1 + 0.963 509 103 792 603 127 658 367 156 982 421 874 851 84;
  • 13) 0.963 509 103 792 603 127 658 367 156 982 421 874 851 84 × 2 = 1 + 0.927 018 207 585 206 255 316 734 313 964 843 749 703 68;
  • 14) 0.927 018 207 585 206 255 316 734 313 964 843 749 703 68 × 2 = 1 + 0.854 036 415 170 412 510 633 468 627 929 687 499 407 36;
  • 15) 0.854 036 415 170 412 510 633 468 627 929 687 499 407 36 × 2 = 1 + 0.708 072 830 340 825 021 266 937 255 859 374 998 814 72;
  • 16) 0.708 072 830 340 825 021 266 937 255 859 374 998 814 72 × 2 = 1 + 0.416 145 660 681 650 042 533 874 511 718 749 997 629 44;
  • 17) 0.416 145 660 681 650 042 533 874 511 718 749 997 629 44 × 2 = 0 + 0.832 291 321 363 300 085 067 749 023 437 499 995 258 88;
  • 18) 0.832 291 321 363 300 085 067 749 023 437 499 995 258 88 × 2 = 1 + 0.664 582 642 726 600 170 135 498 046 874 999 990 517 76;
  • 19) 0.664 582 642 726 600 170 135 498 046 874 999 990 517 76 × 2 = 1 + 0.329 165 285 453 200 340 270 996 093 749 999 981 035 52;
  • 20) 0.329 165 285 453 200 340 270 996 093 749 999 981 035 52 × 2 = 0 + 0.658 330 570 906 400 680 541 992 187 499 999 962 071 04;
  • 21) 0.658 330 570 906 400 680 541 992 187 499 999 962 071 04 × 2 = 1 + 0.316 661 141 812 801 361 083 984 374 999 999 924 142 08;
  • 22) 0.316 661 141 812 801 361 083 984 374 999 999 924 142 08 × 2 = 0 + 0.633 322 283 625 602 722 167 968 749 999 999 848 284 16;
  • 23) 0.633 322 283 625 602 722 167 968 749 999 999 848 284 16 × 2 = 1 + 0.266 644 567 251 205 444 335 937 499 999 999 696 568 32;
  • 24) 0.266 644 567 251 205 444 335 937 499 999 999 696 568 32 × 2 = 0 + 0.533 289 134 502 410 888 671 874 999 999 999 393 136 64;
  • 25) 0.533 289 134 502 410 888 671 874 999 999 999 393 136 64 × 2 = 1 + 0.066 578 269 004 821 777 343 749 999 999 998 786 273 28;
  • 26) 0.066 578 269 004 821 777 343 749 999 999 998 786 273 28 × 2 = 0 + 0.133 156 538 009 643 554 687 499 999 999 997 572 546 56;
  • 27) 0.133 156 538 009 643 554 687 499 999 999 997 572 546 56 × 2 = 0 + 0.266 313 076 019 287 109 374 999 999 999 995 145 093 12;
  • 28) 0.266 313 076 019 287 109 374 999 999 999 995 145 093 12 × 2 = 0 + 0.532 626 152 038 574 218 749 999 999 999 990 290 186 24;
  • 29) 0.532 626 152 038 574 218 749 999 999 999 990 290 186 24 × 2 = 1 + 0.065 252 304 077 148 437 499 999 999 999 980 580 372 48;
  • 30) 0.065 252 304 077 148 437 499 999 999 999 980 580 372 48 × 2 = 0 + 0.130 504 608 154 296 874 999 999 999 999 961 160 744 96;
  • 31) 0.130 504 608 154 296 874 999 999 999 999 961 160 744 96 × 2 = 0 + 0.261 009 216 308 593 749 999 999 999 999 922 321 489 92;
  • 32) 0.261 009 216 308 593 749 999 999 999 999 922 321 489 92 × 2 = 0 + 0.522 018 432 617 187 499 999 999 999 999 844 642 979 84;
  • 33) 0.522 018 432 617 187 499 999 999 999 999 844 642 979 84 × 2 = 1 + 0.044 036 865 234 374 999 999 999 999 999 689 285 959 68;
  • 34) 0.044 036 865 234 374 999 999 999 999 999 689 285 959 68 × 2 = 0 + 0.088 073 730 468 749 999 999 999 999 999 378 571 919 36;
  • 35) 0.088 073 730 468 749 999 999 999 999 999 378 571 919 36 × 2 = 0 + 0.176 147 460 937 499 999 999 999 999 998 757 143 838 72;
  • 36) 0.176 147 460 937 499 999 999 999 999 998 757 143 838 72 × 2 = 0 + 0.352 294 921 874 999 999 999 999 999 997 514 287 677 44;
  • 37) 0.352 294 921 874 999 999 999 999 999 997 514 287 677 44 × 2 = 0 + 0.704 589 843 749 999 999 999 999 999 995 028 575 354 88;
  • 38) 0.704 589 843 749 999 999 999 999 999 995 028 575 354 88 × 2 = 1 + 0.409 179 687 499 999 999 999 999 999 990 057 150 709 76;
  • 39) 0.409 179 687 499 999 999 999 999 999 990 057 150 709 76 × 2 = 0 + 0.818 359 374 999 999 999 999 999 999 980 114 301 419 52;
  • 40) 0.818 359 374 999 999 999 999 999 999 980 114 301 419 52 × 2 = 1 + 0.636 718 749 999 999 999 999 999 999 960 228 602 839 04;
  • 41) 0.636 718 749 999 999 999 999 999 999 960 228 602 839 04 × 2 = 1 + 0.273 437 499 999 999 999 999 999 999 920 457 205 678 08;
  • 42) 0.273 437 499 999 999 999 999 999 999 920 457 205 678 08 × 2 = 0 + 0.546 874 999 999 999 999 999 999 999 840 914 411 356 16;
  • 43) 0.546 874 999 999 999 999 999 999 999 840 914 411 356 16 × 2 = 1 + 0.093 749 999 999 999 999 999 999 999 681 828 822 712 32;
  • 44) 0.093 749 999 999 999 999 999 999 999 681 828 822 712 32 × 2 = 0 + 0.187 499 999 999 999 999 999 999 999 363 657 645 424 64;
  • 45) 0.187 499 999 999 999 999 999 999 999 363 657 645 424 64 × 2 = 0 + 0.374 999 999 999 999 999 999 999 998 727 315 290 849 28;
  • 46) 0.374 999 999 999 999 999 999 999 998 727 315 290 849 28 × 2 = 0 + 0.749 999 999 999 999 999 999 999 997 454 630 581 698 56;
  • 47) 0.749 999 999 999 999 999 999 999 997 454 630 581 698 56 × 2 = 1 + 0.499 999 999 999 999 999 999 999 994 909 261 163 397 12;
  • 48) 0.499 999 999 999 999 999 999 999 994 909 261 163 397 12 × 2 = 0 + 0.999 999 999 999 999 999 999 999 989 818 522 326 794 24;
  • 49) 0.999 999 999 999 999 999 999 999 989 818 522 326 794 24 × 2 = 1 + 0.999 999 999 999 999 999 999 999 979 637 044 653 588 48;
  • 50) 0.999 999 999 999 999 999 999 999 979 637 044 653 588 48 × 2 = 1 + 0.999 999 999 999 999 999 999 999 959 274 089 307 176 96;
  • 51) 0.999 999 999 999 999 999 999 999 959 274 089 307 176 96 × 2 = 1 + 0.999 999 999 999 999 999 999 999 918 548 178 614 353 92;
  • 52) 0.999 999 999 999 999 999 999 999 918 548 178 614 353 92 × 2 = 1 + 0.999 999 999 999 999 999 999 999 837 096 357 228 707 84;
  • 53) 0.999 999 999 999 999 999 999 999 837 096 357 228 707 84 × 2 = 1 + 0.999 999 999 999 999 999 999 999 674 192 714 457 415 68;

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...)


4. 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:


0.141 592 653 589 793 115 997 963 468 544 185 161 590 54(10) =

0.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0010 1111 1(2)


5. Positive number before normalization:

3.141 592 653 589 793 115 997 963 468 544 185 161 590 54(10) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0010 1111 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


3.141 592 653 589 793 115 997 963 468 544 185 161 590 54(10) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0010 1111 1(2) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1000 0101 1010 0010 1111 1(2) × 20 =

1.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 0111 11(2) × 21


7. 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 0 (a positive number)

Exponent (unadjusted): 1

Mantissa (not normalized):
1.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 0111 11


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

1 + 2(11-1) - 1 =

(1 + 1 023)(10) =

1 024(10)


9. 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;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 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;

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

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


Exponent (adjusted) =

1024(10) =

100 0000 0000(2)


11. 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. 1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 0111 11 =

1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 0111


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

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
100 0000 0000

Mantissa (52 bits) =
1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 0111


The base ten decimal number 3.141 592 653 589 793 115 997 963 468 544 185 161 590 54 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0000 0000 - 1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 0111

More operations with decimal numbers converted to 64 bit double precision IEEE 754 binary floating point representation:

» Number 3.141 592 653 589 793 115 997 963 468 544 185 161 590 53 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

» Number 3.141 592 653 589 793 115 997 963 468 544 185 161 590 55 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

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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