-0.000 000 000 000 176 659 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 659(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
-0.000 000 000 000 176 659(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:

|-0.000 000 000 000 176 659| = 0.000 000 000 000 176 659


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: 0.000 000 000 000 176 659.

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) 0.000 000 000 000 176 659 × 2 = 0 + 0.000 000 000 000 353 318;
  • 2) 0.000 000 000 000 353 318 × 2 = 0 + 0.000 000 000 000 706 636;
  • 3) 0.000 000 000 000 706 636 × 2 = 0 + 0.000 000 000 001 413 272;
  • 4) 0.000 000 000 001 413 272 × 2 = 0 + 0.000 000 000 002 826 544;
  • 5) 0.000 000 000 002 826 544 × 2 = 0 + 0.000 000 000 005 653 088;
  • 6) 0.000 000 000 005 653 088 × 2 = 0 + 0.000 000 000 011 306 176;
  • 7) 0.000 000 000 011 306 176 × 2 = 0 + 0.000 000 000 022 612 352;
  • 8) 0.000 000 000 022 612 352 × 2 = 0 + 0.000 000 000 045 224 704;
  • 9) 0.000 000 000 045 224 704 × 2 = 0 + 0.000 000 000 090 449 408;
  • 10) 0.000 000 000 090 449 408 × 2 = 0 + 0.000 000 000 180 898 816;
  • 11) 0.000 000 000 180 898 816 × 2 = 0 + 0.000 000 000 361 797 632;
  • 12) 0.000 000 000 361 797 632 × 2 = 0 + 0.000 000 000 723 595 264;
  • 13) 0.000 000 000 723 595 264 × 2 = 0 + 0.000 000 001 447 190 528;
  • 14) 0.000 000 001 447 190 528 × 2 = 0 + 0.000 000 002 894 381 056;
  • 15) 0.000 000 002 894 381 056 × 2 = 0 + 0.000 000 005 788 762 112;
  • 16) 0.000 000 005 788 762 112 × 2 = 0 + 0.000 000 011 577 524 224;
  • 17) 0.000 000 011 577 524 224 × 2 = 0 + 0.000 000 023 155 048 448;
  • 18) 0.000 000 023 155 048 448 × 2 = 0 + 0.000 000 046 310 096 896;
  • 19) 0.000 000 046 310 096 896 × 2 = 0 + 0.000 000 092 620 193 792;
  • 20) 0.000 000 092 620 193 792 × 2 = 0 + 0.000 000 185 240 387 584;
  • 21) 0.000 000 185 240 387 584 × 2 = 0 + 0.000 000 370 480 775 168;
  • 22) 0.000 000 370 480 775 168 × 2 = 0 + 0.000 000 740 961 550 336;
  • 23) 0.000 000 740 961 550 336 × 2 = 0 + 0.000 001 481 923 100 672;
  • 24) 0.000 001 481 923 100 672 × 2 = 0 + 0.000 002 963 846 201 344;
  • 25) 0.000 002 963 846 201 344 × 2 = 0 + 0.000 005 927 692 402 688;
  • 26) 0.000 005 927 692 402 688 × 2 = 0 + 0.000 011 855 384 805 376;
  • 27) 0.000 011 855 384 805 376 × 2 = 0 + 0.000 023 710 769 610 752;
  • 28) 0.000 023 710 769 610 752 × 2 = 0 + 0.000 047 421 539 221 504;
  • 29) 0.000 047 421 539 221 504 × 2 = 0 + 0.000 094 843 078 443 008;
  • 30) 0.000 094 843 078 443 008 × 2 = 0 + 0.000 189 686 156 886 016;
  • 31) 0.000 189 686 156 886 016 × 2 = 0 + 0.000 379 372 313 772 032;
  • 32) 0.000 379 372 313 772 032 × 2 = 0 + 0.000 758 744 627 544 064;
  • 33) 0.000 758 744 627 544 064 × 2 = 0 + 0.001 517 489 255 088 128;
  • 34) 0.001 517 489 255 088 128 × 2 = 0 + 0.003 034 978 510 176 256;
  • 35) 0.003 034 978 510 176 256 × 2 = 0 + 0.006 069 957 020 352 512;
  • 36) 0.006 069 957 020 352 512 × 2 = 0 + 0.012 139 914 040 705 024;
  • 37) 0.012 139 914 040 705 024 × 2 = 0 + 0.024 279 828 081 410 048;
  • 38) 0.024 279 828 081 410 048 × 2 = 0 + 0.048 559 656 162 820 096;
  • 39) 0.048 559 656 162 820 096 × 2 = 0 + 0.097 119 312 325 640 192;
  • 40) 0.097 119 312 325 640 192 × 2 = 0 + 0.194 238 624 651 280 384;
  • 41) 0.194 238 624 651 280 384 × 2 = 0 + 0.388 477 249 302 560 768;
  • 42) 0.388 477 249 302 560 768 × 2 = 0 + 0.776 954 498 605 121 536;
  • 43) 0.776 954 498 605 121 536 × 2 = 1 + 0.553 908 997 210 243 072;
  • 44) 0.553 908 997 210 243 072 × 2 = 1 + 0.107 817 994 420 486 144;
  • 45) 0.107 817 994 420 486 144 × 2 = 0 + 0.215 635 988 840 972 288;
  • 46) 0.215 635 988 840 972 288 × 2 = 0 + 0.431 271 977 681 944 576;
  • 47) 0.431 271 977 681 944 576 × 2 = 0 + 0.862 543 955 363 889 152;
  • 48) 0.862 543 955 363 889 152 × 2 = 1 + 0.725 087 910 727 778 304;
  • 49) 0.725 087 910 727 778 304 × 2 = 1 + 0.450 175 821 455 556 608;
  • 50) 0.450 175 821 455 556 608 × 2 = 0 + 0.900 351 642 911 113 216;
  • 51) 0.900 351 642 911 113 216 × 2 = 1 + 0.800 703 285 822 226 432;
  • 52) 0.800 703 285 822 226 432 × 2 = 1 + 0.601 406 571 644 452 864;
  • 53) 0.601 406 571 644 452 864 × 2 = 1 + 0.202 813 143 288 905 728;
  • 54) 0.202 813 143 288 905 728 × 2 = 0 + 0.405 626 286 577 811 456;
  • 55) 0.405 626 286 577 811 456 × 2 = 0 + 0.811 252 573 155 622 912;
  • 56) 0.811 252 573 155 622 912 × 2 = 1 + 0.622 505 146 311 245 824;
  • 57) 0.622 505 146 311 245 824 × 2 = 1 + 0.245 010 292 622 491 648;
  • 58) 0.245 010 292 622 491 648 × 2 = 0 + 0.490 020 585 244 983 296;
  • 59) 0.490 020 585 244 983 296 × 2 = 0 + 0.980 041 170 489 966 592;
  • 60) 0.980 041 170 489 966 592 × 2 = 1 + 0.960 082 340 979 933 184;
  • 61) 0.960 082 340 979 933 184 × 2 = 1 + 0.920 164 681 959 866 368;
  • 62) 0.920 164 681 959 866 368 × 2 = 1 + 0.840 329 363 919 732 736;
  • 63) 0.840 329 363 919 732 736 × 2 = 1 + 0.680 658 727 839 465 472;
  • 64) 0.680 658 727 839 465 472 × 2 = 1 + 0.361 317 455 678 930 944;
  • 65) 0.361 317 455 678 930 944 × 2 = 0 + 0.722 634 911 357 861 888;
  • 66) 0.722 634 911 357 861 888 × 2 = 1 + 0.445 269 822 715 723 776;
  • 67) 0.445 269 822 715 723 776 × 2 = 0 + 0.890 539 645 431 447 552;
  • 68) 0.890 539 645 431 447 552 × 2 = 1 + 0.781 079 290 862 895 104;
  • 69) 0.781 079 290 862 895 104 × 2 = 1 + 0.562 158 581 725 790 208;
  • 70) 0.562 158 581 725 790 208 × 2 = 1 + 0.124 317 163 451 580 416;
  • 71) 0.124 317 163 451 580 416 × 2 = 0 + 0.248 634 326 903 160 832;
  • 72) 0.248 634 326 903 160 832 × 2 = 0 + 0.497 268 653 806 321 664;
  • 73) 0.497 268 653 806 321 664 × 2 = 0 + 0.994 537 307 612 643 328;
  • 74) 0.994 537 307 612 643 328 × 2 = 1 + 0.989 074 615 225 286 656;
  • 75) 0.989 074 615 225 286 656 × 2 = 1 + 0.978 149 230 450 573 312;
  • 76) 0.978 149 230 450 573 312 × 2 = 1 + 0.956 298 460 901 146 624;
  • 77) 0.956 298 460 901 146 624 × 2 = 1 + 0.912 596 921 802 293 248;
  • 78) 0.912 596 921 802 293 248 × 2 = 1 + 0.825 193 843 604 586 496;
  • 79) 0.825 193 843 604 586 496 × 2 = 1 + 0.650 387 687 209 172 992;
  • 80) 0.650 387 687 209 172 992 × 2 = 1 + 0.300 775 374 418 345 984;
  • 81) 0.300 775 374 418 345 984 × 2 = 0 + 0.601 550 748 836 691 968;
  • 82) 0.601 550 748 836 691 968 × 2 = 1 + 0.203 101 497 673 383 936;
  • 83) 0.203 101 497 673 383 936 × 2 = 0 + 0.406 202 995 346 767 872;
  • 84) 0.406 202 995 346 767 872 × 2 = 0 + 0.812 405 990 693 535 744;
  • 85) 0.812 405 990 693 535 744 × 2 = 1 + 0.624 811 981 387 071 488;
  • 86) 0.624 811 981 387 071 488 × 2 = 1 + 0.249 623 962 774 142 976;
  • 87) 0.249 623 962 774 142 976 × 2 = 0 + 0.499 247 925 548 285 952;
  • 88) 0.499 247 925 548 285 952 × 2 = 0 + 0.998 495 851 096 571 904;
  • 89) 0.998 495 851 096 571 904 × 2 = 1 + 0.996 991 702 193 143 808;
  • 90) 0.996 991 702 193 143 808 × 2 = 1 + 0.993 983 404 386 287 616;
  • 91) 0.993 983 404 386 287 616 × 2 = 1 + 0.987 966 808 772 575 232;
  • 92) 0.987 966 808 772 575 232 × 2 = 1 + 0.975 933 617 545 150 464;
  • 93) 0.975 933 617 545 150 464 × 2 = 1 + 0.951 867 235 090 300 928;
  • 94) 0.951 867 235 090 300 928 × 2 = 1 + 0.903 734 470 180 601 856;
  • 95) 0.903 734 470 180 601 856 × 2 = 1 + 0.807 468 940 361 203 712;

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:


0.000 000 000 000 176 659(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 1001 1001 1111 0101 1100 0111 1111 0100 1100 1111 111(2)

6. Positive number before normalization:

0.000 000 000 000 176 659(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 1001 1001 1111 0101 1100 0111 1111 0100 1100 1111 111(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 43 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 176 659(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 1001 1001 1111 0101 1100 0111 1111 0100 1100 1111 111(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 1001 1001 1111 0101 1100 0111 1111 0100 1100 1111 111(2) × 20 =

1.1000 1101 1100 1100 1111 1010 1110 0011 1111 1010 0110 0111 1111(2) × 2-43


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): -43

Mantissa (not normalized):
1.1000 1101 1100 1100 1111 1010 1110 0011 1111 1010 0110 0111 1111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-43 + 2(11-1) - 1 =

(-43 + 1 023)(10) =

980(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;
  • 980 ÷ 2 = 490 + 0;
  • 490 ÷ 2 = 245 + 0;
  • 245 ÷ 2 = 122 + 1;
  • 122 ÷ 2 = 61 + 0;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 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) =

980(10) =

011 1101 0100(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. 1000 1101 1100 1100 1111 1010 1110 0011 1111 1010 0110 0111 1111 =

1000 1101 1100 1100 1111 1010 1110 0011 1111 1010 0110 0111 1111


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 1101 0100

Mantissa (52 bits) =
1000 1101 1100 1100 1111 1010 1110 0011 1111 1010 0110 0111 1111


Decimal number -0.000 000 000 000 176 659 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1100 1100 1111 1010 1110 0011 1111 1010 0110 0111 1111

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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