-0.000 000 349 613 904 800 000 252 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 349 613 904 800 000 252 4(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 349 613 904 800 000 252 4(10) to 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:

|-0.000 000 349 613 904 800 000 252 4| = 0.000 000 349 613 904 800 000 252 4


2. First, convert to binary (in base 2) the integer part: 0.
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;
  • 0 ÷ 2 = 0 + 0;

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.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 349 613 904 800 000 252 4.

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) 0.000 000 349 613 904 800 000 252 4 × 2 = 0 + 0.000 000 699 227 809 600 000 504 8;
  • 2) 0.000 000 699 227 809 600 000 504 8 × 2 = 0 + 0.000 001 398 455 619 200 001 009 6;
  • 3) 0.000 001 398 455 619 200 001 009 6 × 2 = 0 + 0.000 002 796 911 238 400 002 019 2;
  • 4) 0.000 002 796 911 238 400 002 019 2 × 2 = 0 + 0.000 005 593 822 476 800 004 038 4;
  • 5) 0.000 005 593 822 476 800 004 038 4 × 2 = 0 + 0.000 011 187 644 953 600 008 076 8;
  • 6) 0.000 011 187 644 953 600 008 076 8 × 2 = 0 + 0.000 022 375 289 907 200 016 153 6;
  • 7) 0.000 022 375 289 907 200 016 153 6 × 2 = 0 + 0.000 044 750 579 814 400 032 307 2;
  • 8) 0.000 044 750 579 814 400 032 307 2 × 2 = 0 + 0.000 089 501 159 628 800 064 614 4;
  • 9) 0.000 089 501 159 628 800 064 614 4 × 2 = 0 + 0.000 179 002 319 257 600 129 228 8;
  • 10) 0.000 179 002 319 257 600 129 228 8 × 2 = 0 + 0.000 358 004 638 515 200 258 457 6;
  • 11) 0.000 358 004 638 515 200 258 457 6 × 2 = 0 + 0.000 716 009 277 030 400 516 915 2;
  • 12) 0.000 716 009 277 030 400 516 915 2 × 2 = 0 + 0.001 432 018 554 060 801 033 830 4;
  • 13) 0.001 432 018 554 060 801 033 830 4 × 2 = 0 + 0.002 864 037 108 121 602 067 660 8;
  • 14) 0.002 864 037 108 121 602 067 660 8 × 2 = 0 + 0.005 728 074 216 243 204 135 321 6;
  • 15) 0.005 728 074 216 243 204 135 321 6 × 2 = 0 + 0.011 456 148 432 486 408 270 643 2;
  • 16) 0.011 456 148 432 486 408 270 643 2 × 2 = 0 + 0.022 912 296 864 972 816 541 286 4;
  • 17) 0.022 912 296 864 972 816 541 286 4 × 2 = 0 + 0.045 824 593 729 945 633 082 572 8;
  • 18) 0.045 824 593 729 945 633 082 572 8 × 2 = 0 + 0.091 649 187 459 891 266 165 145 6;
  • 19) 0.091 649 187 459 891 266 165 145 6 × 2 = 0 + 0.183 298 374 919 782 532 330 291 2;
  • 20) 0.183 298 374 919 782 532 330 291 2 × 2 = 0 + 0.366 596 749 839 565 064 660 582 4;
  • 21) 0.366 596 749 839 565 064 660 582 4 × 2 = 0 + 0.733 193 499 679 130 129 321 164 8;
  • 22) 0.733 193 499 679 130 129 321 164 8 × 2 = 1 + 0.466 386 999 358 260 258 642 329 6;
  • 23) 0.466 386 999 358 260 258 642 329 6 × 2 = 0 + 0.932 773 998 716 520 517 284 659 2;
  • 24) 0.932 773 998 716 520 517 284 659 2 × 2 = 1 + 0.865 547 997 433 041 034 569 318 4;
  • 25) 0.865 547 997 433 041 034 569 318 4 × 2 = 1 + 0.731 095 994 866 082 069 138 636 8;
  • 26) 0.731 095 994 866 082 069 138 636 8 × 2 = 1 + 0.462 191 989 732 164 138 277 273 6;
  • 27) 0.462 191 989 732 164 138 277 273 6 × 2 = 0 + 0.924 383 979 464 328 276 554 547 2;
  • 28) 0.924 383 979 464 328 276 554 547 2 × 2 = 1 + 0.848 767 958 928 656 553 109 094 4;
  • 29) 0.848 767 958 928 656 553 109 094 4 × 2 = 1 + 0.697 535 917 857 313 106 218 188 8;
  • 30) 0.697 535 917 857 313 106 218 188 8 × 2 = 1 + 0.395 071 835 714 626 212 436 377 6;
  • 31) 0.395 071 835 714 626 212 436 377 6 × 2 = 0 + 0.790 143 671 429 252 424 872 755 2;
  • 32) 0.790 143 671 429 252 424 872 755 2 × 2 = 1 + 0.580 287 342 858 504 849 745 510 4;
  • 33) 0.580 287 342 858 504 849 745 510 4 × 2 = 1 + 0.160 574 685 717 009 699 491 020 8;
  • 34) 0.160 574 685 717 009 699 491 020 8 × 2 = 0 + 0.321 149 371 434 019 398 982 041 6;
  • 35) 0.321 149 371 434 019 398 982 041 6 × 2 = 0 + 0.642 298 742 868 038 797 964 083 2;
  • 36) 0.642 298 742 868 038 797 964 083 2 × 2 = 1 + 0.284 597 485 736 077 595 928 166 4;
  • 37) 0.284 597 485 736 077 595 928 166 4 × 2 = 0 + 0.569 194 971 472 155 191 856 332 8;
  • 38) 0.569 194 971 472 155 191 856 332 8 × 2 = 1 + 0.138 389 942 944 310 383 712 665 6;
  • 39) 0.138 389 942 944 310 383 712 665 6 × 2 = 0 + 0.276 779 885 888 620 767 425 331 2;
  • 40) 0.276 779 885 888 620 767 425 331 2 × 2 = 0 + 0.553 559 771 777 241 534 850 662 4;
  • 41) 0.553 559 771 777 241 534 850 662 4 × 2 = 1 + 0.107 119 543 554 483 069 701 324 8;
  • 42) 0.107 119 543 554 483 069 701 324 8 × 2 = 0 + 0.214 239 087 108 966 139 402 649 6;
  • 43) 0.214 239 087 108 966 139 402 649 6 × 2 = 0 + 0.428 478 174 217 932 278 805 299 2;
  • 44) 0.428 478 174 217 932 278 805 299 2 × 2 = 0 + 0.856 956 348 435 864 557 610 598 4;
  • 45) 0.856 956 348 435 864 557 610 598 4 × 2 = 1 + 0.713 912 696 871 729 115 221 196 8;
  • 46) 0.713 912 696 871 729 115 221 196 8 × 2 = 1 + 0.427 825 393 743 458 230 442 393 6;
  • 47) 0.427 825 393 743 458 230 442 393 6 × 2 = 0 + 0.855 650 787 486 916 460 884 787 2;
  • 48) 0.855 650 787 486 916 460 884 787 2 × 2 = 1 + 0.711 301 574 973 832 921 769 574 4;
  • 49) 0.711 301 574 973 832 921 769 574 4 × 2 = 1 + 0.422 603 149 947 665 843 539 148 8;
  • 50) 0.422 603 149 947 665 843 539 148 8 × 2 = 0 + 0.845 206 299 895 331 687 078 297 6;
  • 51) 0.845 206 299 895 331 687 078 297 6 × 2 = 1 + 0.690 412 599 790 663 374 156 595 2;
  • 52) 0.690 412 599 790 663 374 156 595 2 × 2 = 1 + 0.380 825 199 581 326 748 313 190 4;
  • 53) 0.380 825 199 581 326 748 313 190 4 × 2 = 0 + 0.761 650 399 162 653 496 626 380 8;
  • 54) 0.761 650 399 162 653 496 626 380 8 × 2 = 1 + 0.523 300 798 325 306 993 252 761 6;
  • 55) 0.523 300 798 325 306 993 252 761 6 × 2 = 1 + 0.046 601 596 650 613 986 505 523 2;
  • 56) 0.046 601 596 650 613 986 505 523 2 × 2 = 0 + 0.093 203 193 301 227 973 011 046 4;
  • 57) 0.093 203 193 301 227 973 011 046 4 × 2 = 0 + 0.186 406 386 602 455 946 022 092 8;
  • 58) 0.186 406 386 602 455 946 022 092 8 × 2 = 0 + 0.372 812 773 204 911 892 044 185 6;
  • 59) 0.372 812 773 204 911 892 044 185 6 × 2 = 0 + 0.745 625 546 409 823 784 088 371 2;
  • 60) 0.745 625 546 409 823 784 088 371 2 × 2 = 1 + 0.491 251 092 819 647 568 176 742 4;
  • 61) 0.491 251 092 819 647 568 176 742 4 × 2 = 0 + 0.982 502 185 639 295 136 353 484 8;
  • 62) 0.982 502 185 639 295 136 353 484 8 × 2 = 1 + 0.965 004 371 278 590 272 706 969 6;
  • 63) 0.965 004 371 278 590 272 706 969 6 × 2 = 1 + 0.930 008 742 557 180 545 413 939 2;
  • 64) 0.930 008 742 557 180 545 413 939 2 × 2 = 1 + 0.860 017 485 114 361 090 827 878 4;
  • 65) 0.860 017 485 114 361 090 827 878 4 × 2 = 1 + 0.720 034 970 228 722 181 655 756 8;
  • 66) 0.720 034 970 228 722 181 655 756 8 × 2 = 1 + 0.440 069 940 457 444 363 311 513 6;
  • 67) 0.440 069 940 457 444 363 311 513 6 × 2 = 0 + 0.880 139 880 914 888 726 623 027 2;
  • 68) 0.880 139 880 914 888 726 623 027 2 × 2 = 1 + 0.760 279 761 829 777 453 246 054 4;
  • 69) 0.760 279 761 829 777 453 246 054 4 × 2 = 1 + 0.520 559 523 659 554 906 492 108 8;
  • 70) 0.520 559 523 659 554 906 492 108 8 × 2 = 1 + 0.041 119 047 319 109 812 984 217 6;
  • 71) 0.041 119 047 319 109 812 984 217 6 × 2 = 0 + 0.082 238 094 638 219 625 968 435 2;
  • 72) 0.082 238 094 638 219 625 968 435 2 × 2 = 0 + 0.164 476 189 276 439 251 936 870 4;
  • 73) 0.164 476 189 276 439 251 936 870 4 × 2 = 0 + 0.328 952 378 552 878 503 873 740 8;
  • 74) 0.328 952 378 552 878 503 873 740 8 × 2 = 0 + 0.657 904 757 105 757 007 747 481 6;

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 - the converted number we get in the end will be just a very good approximation of the initial one).


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:


0.000 000 349 613 904 800 000 252 4(10) =

0.0000 0000 0000 0000 0000 0101 1101 1101 1001 0100 1000 1101 1011 0110 0001 0111 1101 1100 00(2)

6. Positive number before normalization:

0.000 000 349 613 904 800 000 252 4(10) =

0.0000 0000 0000 0000 0000 0101 1101 1101 1001 0100 1000 1101 1011 0110 0001 0111 1101 1100 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 22 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 349 613 904 800 000 252 4(10) =

0.0000 0000 0000 0000 0000 0101 1101 1101 1001 0100 1000 1101 1011 0110 0001 0111 1101 1100 00(2) =

0.0000 0000 0000 0000 0000 0101 1101 1101 1001 0100 1000 1101 1011 0110 0001 0111 1101 1100 00(2) × 20 =

1.0111 0111 0110 0101 0010 0011 0110 1101 1000 0101 1111 0111 0000(2) × 2-22


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): -22

Mantissa (not normalized):
1.0111 0111 0110 0101 0010 0011 0110 1101 1000 0101 1111 0111 0000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-22 + 2(11-1) - 1 =

(-22 + 1 023)(10) =

1 001(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;
  • 1 001 ÷ 2 = 500 + 1;
  • 500 ÷ 2 = 250 + 0;
  • 250 ÷ 2 = 125 + 0;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

1001(10) =

011 1110 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, only if necessary (not the case here).


Mantissa (normalized) =

1. 0111 0111 0110 0101 0010 0011 0110 1101 1000 0101 1111 0111 0000 =

0111 0111 0110 0101 0010 0011 0110 1101 1000 0101 1111 0111 0000


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) =
011 1110 1001

Mantissa (52 bits) =
0111 0111 0110 0101 0010 0011 0110 1101 1000 0101 1111 0111 0000


Decimal number -0.000 000 349 613 904 800 000 252 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1001 - 0111 0111 0110 0101 0010 0011 0110 1101 1000 0101 1111 0111 0000

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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