64bit IEEE 754: Decimal -> Double Precision Floating Point Binary: 1 000 001 000 100 000 000 000 000 000 014 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 1 000 001 000 100 000 000 000 000 000 014(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. 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;
  • 1 000 001 000 100 000 000 000 000 000 014 ÷ 2 = 500 000 500 050 000 000 000 000 000 007 + 0;
  • 500 000 500 050 000 000 000 000 000 007 ÷ 2 = 250 000 250 025 000 000 000 000 000 003 + 1;
  • 250 000 250 025 000 000 000 000 000 003 ÷ 2 = 125 000 125 012 500 000 000 000 000 001 + 1;
  • 125 000 125 012 500 000 000 000 000 001 ÷ 2 = 62 500 062 506 250 000 000 000 000 000 + 1;
  • 62 500 062 506 250 000 000 000 000 000 ÷ 2 = 31 250 031 253 125 000 000 000 000 000 + 0;
  • 31 250 031 253 125 000 000 000 000 000 ÷ 2 = 15 625 015 626 562 500 000 000 000 000 + 0;
  • 15 625 015 626 562 500 000 000 000 000 ÷ 2 = 7 812 507 813 281 250 000 000 000 000 + 0;
  • 7 812 507 813 281 250 000 000 000 000 ÷ 2 = 3 906 253 906 640 625 000 000 000 000 + 0;
  • 3 906 253 906 640 625 000 000 000 000 ÷ 2 = 1 953 126 953 320 312 500 000 000 000 + 0;
  • 1 953 126 953 320 312 500 000 000 000 ÷ 2 = 976 563 476 660 156 250 000 000 000 + 0;
  • 976 563 476 660 156 250 000 000 000 ÷ 2 = 488 281 738 330 078 125 000 000 000 + 0;
  • 488 281 738 330 078 125 000 000 000 ÷ 2 = 244 140 869 165 039 062 500 000 000 + 0;
  • 244 140 869 165 039 062 500 000 000 ÷ 2 = 122 070 434 582 519 531 250 000 000 + 0;
  • 122 070 434 582 519 531 250 000 000 ÷ 2 = 61 035 217 291 259 765 625 000 000 + 0;
  • 61 035 217 291 259 765 625 000 000 ÷ 2 = 30 517 608 645 629 882 812 500 000 + 0;
  • 30 517 608 645 629 882 812 500 000 ÷ 2 = 15 258 804 322 814 941 406 250 000 + 0;
  • 15 258 804 322 814 941 406 250 000 ÷ 2 = 7 629 402 161 407 470 703 125 000 + 0;
  • 7 629 402 161 407 470 703 125 000 ÷ 2 = 3 814 701 080 703 735 351 562 500 + 0;
  • 3 814 701 080 703 735 351 562 500 ÷ 2 = 1 907 350 540 351 867 675 781 250 + 0;
  • 1 907 350 540 351 867 675 781 250 ÷ 2 = 953 675 270 175 933 837 890 625 + 0;
  • 953 675 270 175 933 837 890 625 ÷ 2 = 476 837 635 087 966 918 945 312 + 1;
  • 476 837 635 087 966 918 945 312 ÷ 2 = 238 418 817 543 983 459 472 656 + 0;
  • 238 418 817 543 983 459 472 656 ÷ 2 = 119 209 408 771 991 729 736 328 + 0;
  • 119 209 408 771 991 729 736 328 ÷ 2 = 59 604 704 385 995 864 868 164 + 0;
  • 59 604 704 385 995 864 868 164 ÷ 2 = 29 802 352 192 997 932 434 082 + 0;
  • 29 802 352 192 997 932 434 082 ÷ 2 = 14 901 176 096 498 966 217 041 + 0;
  • 14 901 176 096 498 966 217 041 ÷ 2 = 7 450 588 048 249 483 108 520 + 1;
  • 7 450 588 048 249 483 108 520 ÷ 2 = 3 725 294 024 124 741 554 260 + 0;
  • 3 725 294 024 124 741 554 260 ÷ 2 = 1 862 647 012 062 370 777 130 + 0;
  • 1 862 647 012 062 370 777 130 ÷ 2 = 931 323 506 031 185 388 565 + 0;
  • 931 323 506 031 185 388 565 ÷ 2 = 465 661 753 015 592 694 282 + 1;
  • 465 661 753 015 592 694 282 ÷ 2 = 232 830 876 507 796 347 141 + 0;
  • 232 830 876 507 796 347 141 ÷ 2 = 116 415 438 253 898 173 570 + 1;
  • 116 415 438 253 898 173 570 ÷ 2 = 58 207 719 126 949 086 785 + 0;
  • 58 207 719 126 949 086 785 ÷ 2 = 29 103 859 563 474 543 392 + 1;
  • 29 103 859 563 474 543 392 ÷ 2 = 14 551 929 781 737 271 696 + 0;
  • 14 551 929 781 737 271 696 ÷ 2 = 7 275 964 890 868 635 848 + 0;
  • 7 275 964 890 868 635 848 ÷ 2 = 3 637 982 445 434 317 924 + 0;
  • 3 637 982 445 434 317 924 ÷ 2 = 1 818 991 222 717 158 962 + 0;
  • 1 818 991 222 717 158 962 ÷ 2 = 909 495 611 358 579 481 + 0;
  • 909 495 611 358 579 481 ÷ 2 = 454 747 805 679 289 740 + 1;
  • 454 747 805 679 289 740 ÷ 2 = 227 373 902 839 644 870 + 0;
  • 227 373 902 839 644 870 ÷ 2 = 113 686 951 419 822 435 + 0;
  • 113 686 951 419 822 435 ÷ 2 = 56 843 475 709 911 217 + 1;
  • 56 843 475 709 911 217 ÷ 2 = 28 421 737 854 955 608 + 1;
  • 28 421 737 854 955 608 ÷ 2 = 14 210 868 927 477 804 + 0;
  • 14 210 868 927 477 804 ÷ 2 = 7 105 434 463 738 902 + 0;
  • 7 105 434 463 738 902 ÷ 2 = 3 552 717 231 869 451 + 0;
  • 3 552 717 231 869 451 ÷ 2 = 1 776 358 615 934 725 + 1;
  • 1 776 358 615 934 725 ÷ 2 = 888 179 307 967 362 + 1;
  • 888 179 307 967 362 ÷ 2 = 444 089 653 983 681 + 0;
  • 444 089 653 983 681 ÷ 2 = 222 044 826 991 840 + 1;
  • 222 044 826 991 840 ÷ 2 = 111 022 413 495 920 + 0;
  • 111 022 413 495 920 ÷ 2 = 55 511 206 747 960 + 0;
  • 55 511 206 747 960 ÷ 2 = 27 755 603 373 980 + 0;
  • 27 755 603 373 980 ÷ 2 = 13 877 801 686 990 + 0;
  • 13 877 801 686 990 ÷ 2 = 6 938 900 843 495 + 0;
  • 6 938 900 843 495 ÷ 2 = 3 469 450 421 747 + 1;
  • 3 469 450 421 747 ÷ 2 = 1 734 725 210 873 + 1;
  • 1 734 725 210 873 ÷ 2 = 867 362 605 436 + 1;
  • 867 362 605 436 ÷ 2 = 433 681 302 718 + 0;
  • 433 681 302 718 ÷ 2 = 216 840 651 359 + 0;
  • 216 840 651 359 ÷ 2 = 108 420 325 679 + 1;
  • 108 420 325 679 ÷ 2 = 54 210 162 839 + 1;
  • 54 210 162 839 ÷ 2 = 27 105 081 419 + 1;
  • 27 105 081 419 ÷ 2 = 13 552 540 709 + 1;
  • 13 552 540 709 ÷ 2 = 6 776 270 354 + 1;
  • 6 776 270 354 ÷ 2 = 3 388 135 177 + 0;
  • 3 388 135 177 ÷ 2 = 1 694 067 588 + 1;
  • 1 694 067 588 ÷ 2 = 847 033 794 + 0;
  • 847 033 794 ÷ 2 = 423 516 897 + 0;
  • 423 516 897 ÷ 2 = 211 758 448 + 1;
  • 211 758 448 ÷ 2 = 105 879 224 + 0;
  • 105 879 224 ÷ 2 = 52 939 612 + 0;
  • 52 939 612 ÷ 2 = 26 469 806 + 0;
  • 26 469 806 ÷ 2 = 13 234 903 + 0;
  • 13 234 903 ÷ 2 = 6 617 451 + 1;
  • 6 617 451 ÷ 2 = 3 308 725 + 1;
  • 3 308 725 ÷ 2 = 1 654 362 + 1;
  • 1 654 362 ÷ 2 = 827 181 + 0;
  • 827 181 ÷ 2 = 413 590 + 1;
  • 413 590 ÷ 2 = 206 795 + 0;
  • 206 795 ÷ 2 = 103 397 + 1;
  • 103 397 ÷ 2 = 51 698 + 1;
  • 51 698 ÷ 2 = 25 849 + 0;
  • 25 849 ÷ 2 = 12 924 + 1;
  • 12 924 ÷ 2 = 6 462 + 0;
  • 6 462 ÷ 2 = 3 231 + 0;
  • 3 231 ÷ 2 = 1 615 + 1;
  • 1 615 ÷ 2 = 807 + 1;
  • 807 ÷ 2 = 403 + 1;
  • 403 ÷ 2 = 201 + 1;
  • 201 ÷ 2 = 100 + 1;
  • 100 ÷ 2 = 50 + 0;
  • 50 ÷ 2 = 25 + 0;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the positive number.

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


1 000 001 000 100 000 000 000 000 000 014(10) =


1100 1001 1111 0010 1101 0111 0000 1001 0111 1100 1110 0000 1011 0001 1001 0000 0101 0100 0100 0001 0000 0000 0000 0000 1110(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 99 positions to the left, so that only one non zero digit remains to the left of it:


1 000 001 000 100 000 000 000 000 000 014(10) =


1100 1001 1111 0010 1101 0111 0000 1001 0111 1100 1110 0000 1011 0001 1001 0000 0101 0100 0100 0001 0000 0000 0000 0000 1110(2) =


1100 1001 1111 0010 1101 0111 0000 1001 0111 1100 1110 0000 1011 0001 1001 0000 0101 0100 0100 0001 0000 0000 0000 0000 1110(2) × 20 =


1.1001 0011 1110 0101 1010 1110 0001 0010 1111 1001 1100 0001 0110 0011 0010 0000 1010 1000 1000 0010 0000 0000 0000 0001 110(2) × 299


4. 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): 99


Mantissa (not normalized):
1.1001 0011 1110 0101 1010 1110 0001 0010 1111 1001 1100 0001 0110 0011 0010 0000 1010 1000 1000 0010 0000 0000 0000 0001 110


5. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


99 + 2(11-1) - 1 =


(99 + 1 023)(10) =


1 122(10)


6. 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;
  • 1 122 ÷ 2 = 561 + 0;
  • 561 ÷ 2 = 280 + 1;
  • 280 ÷ 2 = 140 + 0;
  • 140 ÷ 2 = 70 + 0;
  • 70 ÷ 2 = 35 + 0;
  • 35 ÷ 2 = 17 + 1;
  • 17 ÷ 2 = 8 + 1;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =


1122(10) =


100 0110 0010(2)


8. 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 0011 1110 0101 1010 1110 0001 0010 1111 1001 1100 0001 0110 001 1001 0000 0101 0100 0100 0001 0000 0000 0000 0000 1110 =


1001 0011 1110 0101 1010 1110 0001 0010 1111 1001 1100 0001 0110


9. 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 0110 0010


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


The base ten decimal number 1 000 001 000 100 000 000 000 000 000 014 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0110 0010 - 1001 0011 1110 0101 1010 1110 0001 0010 1111 1001 1100 0001 0110

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 1

      62
    • 0

      61
    • 0

      60
    • 0

      59
    • 1

      58
    • 1

      57
    • 0

      56
    • 0

      55
    • 0

      54
    • 1

      53
    • 0

      52
  • Mantissa (52 bits):

    • 1

      51
    • 0

      50
    • 0

      49
    • 1

      48
    • 0

      47
    • 0

      46
    • 1

      45
    • 1

      44
    • 1

      43
    • 1

      42
    • 1

      41
    • 0

      40
    • 0

      39
    • 1

      38
    • 0

      37
    • 1

      36
    • 1

      35
    • 0

      34
    • 1

      33
    • 0

      32
    • 1

      31
    • 1

      30
    • 1

      29
    • 0

      28
    • 0

      27
    • 0

      26
    • 0

      25
    • 1

      24
    • 0

      23
    • 0

      22
    • 1

      21
    • 0

      20
    • 1

      19
    • 1

      18
    • 1

      17
    • 1

      16
    • 1

      15
    • 0

      14
    • 0

      13
    • 1

      12
    • 1

      11
    • 1

      10
    • 0

      9
    • 0

      8
    • 0

      7
    • 0

      6
    • 0

      5
    • 1

      4
    • 0

      3
    • 1

      2
    • 1

      1
    • 0

      0

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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