-0.000 000 000 000 176 563 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 563(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
-0.000 000 000 000 176 563(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:

|-0.000 000 000 000 176 563| = 0.000 000 000 000 176 563


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: 0.000 000 000 000 176 563.

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) 0.000 000 000 000 176 563 × 2 = 0 + 0.000 000 000 000 353 126;
  • 2) 0.000 000 000 000 353 126 × 2 = 0 + 0.000 000 000 000 706 252;
  • 3) 0.000 000 000 000 706 252 × 2 = 0 + 0.000 000 000 001 412 504;
  • 4) 0.000 000 000 001 412 504 × 2 = 0 + 0.000 000 000 002 825 008;
  • 5) 0.000 000 000 002 825 008 × 2 = 0 + 0.000 000 000 005 650 016;
  • 6) 0.000 000 000 005 650 016 × 2 = 0 + 0.000 000 000 011 300 032;
  • 7) 0.000 000 000 011 300 032 × 2 = 0 + 0.000 000 000 022 600 064;
  • 8) 0.000 000 000 022 600 064 × 2 = 0 + 0.000 000 000 045 200 128;
  • 9) 0.000 000 000 045 200 128 × 2 = 0 + 0.000 000 000 090 400 256;
  • 10) 0.000 000 000 090 400 256 × 2 = 0 + 0.000 000 000 180 800 512;
  • 11) 0.000 000 000 180 800 512 × 2 = 0 + 0.000 000 000 361 601 024;
  • 12) 0.000 000 000 361 601 024 × 2 = 0 + 0.000 000 000 723 202 048;
  • 13) 0.000 000 000 723 202 048 × 2 = 0 + 0.000 000 001 446 404 096;
  • 14) 0.000 000 001 446 404 096 × 2 = 0 + 0.000 000 002 892 808 192;
  • 15) 0.000 000 002 892 808 192 × 2 = 0 + 0.000 000 005 785 616 384;
  • 16) 0.000 000 005 785 616 384 × 2 = 0 + 0.000 000 011 571 232 768;
  • 17) 0.000 000 011 571 232 768 × 2 = 0 + 0.000 000 023 142 465 536;
  • 18) 0.000 000 023 142 465 536 × 2 = 0 + 0.000 000 046 284 931 072;
  • 19) 0.000 000 046 284 931 072 × 2 = 0 + 0.000 000 092 569 862 144;
  • 20) 0.000 000 092 569 862 144 × 2 = 0 + 0.000 000 185 139 724 288;
  • 21) 0.000 000 185 139 724 288 × 2 = 0 + 0.000 000 370 279 448 576;
  • 22) 0.000 000 370 279 448 576 × 2 = 0 + 0.000 000 740 558 897 152;
  • 23) 0.000 000 740 558 897 152 × 2 = 0 + 0.000 001 481 117 794 304;
  • 24) 0.000 001 481 117 794 304 × 2 = 0 + 0.000 002 962 235 588 608;
  • 25) 0.000 002 962 235 588 608 × 2 = 0 + 0.000 005 924 471 177 216;
  • 26) 0.000 005 924 471 177 216 × 2 = 0 + 0.000 011 848 942 354 432;
  • 27) 0.000 011 848 942 354 432 × 2 = 0 + 0.000 023 697 884 708 864;
  • 28) 0.000 023 697 884 708 864 × 2 = 0 + 0.000 047 395 769 417 728;
  • 29) 0.000 047 395 769 417 728 × 2 = 0 + 0.000 094 791 538 835 456;
  • 30) 0.000 094 791 538 835 456 × 2 = 0 + 0.000 189 583 077 670 912;
  • 31) 0.000 189 583 077 670 912 × 2 = 0 + 0.000 379 166 155 341 824;
  • 32) 0.000 379 166 155 341 824 × 2 = 0 + 0.000 758 332 310 683 648;
  • 33) 0.000 758 332 310 683 648 × 2 = 0 + 0.001 516 664 621 367 296;
  • 34) 0.001 516 664 621 367 296 × 2 = 0 + 0.003 033 329 242 734 592;
  • 35) 0.003 033 329 242 734 592 × 2 = 0 + 0.006 066 658 485 469 184;
  • 36) 0.006 066 658 485 469 184 × 2 = 0 + 0.012 133 316 970 938 368;
  • 37) 0.012 133 316 970 938 368 × 2 = 0 + 0.024 266 633 941 876 736;
  • 38) 0.024 266 633 941 876 736 × 2 = 0 + 0.048 533 267 883 753 472;
  • 39) 0.048 533 267 883 753 472 × 2 = 0 + 0.097 066 535 767 506 944;
  • 40) 0.097 066 535 767 506 944 × 2 = 0 + 0.194 133 071 535 013 888;
  • 41) 0.194 133 071 535 013 888 × 2 = 0 + 0.388 266 143 070 027 776;
  • 42) 0.388 266 143 070 027 776 × 2 = 0 + 0.776 532 286 140 055 552;
  • 43) 0.776 532 286 140 055 552 × 2 = 1 + 0.553 064 572 280 111 104;
  • 44) 0.553 064 572 280 111 104 × 2 = 1 + 0.106 129 144 560 222 208;
  • 45) 0.106 129 144 560 222 208 × 2 = 0 + 0.212 258 289 120 444 416;
  • 46) 0.212 258 289 120 444 416 × 2 = 0 + 0.424 516 578 240 888 832;
  • 47) 0.424 516 578 240 888 832 × 2 = 0 + 0.849 033 156 481 777 664;
  • 48) 0.849 033 156 481 777 664 × 2 = 1 + 0.698 066 312 963 555 328;
  • 49) 0.698 066 312 963 555 328 × 2 = 1 + 0.396 132 625 927 110 656;
  • 50) 0.396 132 625 927 110 656 × 2 = 0 + 0.792 265 251 854 221 312;
  • 51) 0.792 265 251 854 221 312 × 2 = 1 + 0.584 530 503 708 442 624;
  • 52) 0.584 530 503 708 442 624 × 2 = 1 + 0.169 061 007 416 885 248;
  • 53) 0.169 061 007 416 885 248 × 2 = 0 + 0.338 122 014 833 770 496;
  • 54) 0.338 122 014 833 770 496 × 2 = 0 + 0.676 244 029 667 540 992;
  • 55) 0.676 244 029 667 540 992 × 2 = 1 + 0.352 488 059 335 081 984;
  • 56) 0.352 488 059 335 081 984 × 2 = 0 + 0.704 976 118 670 163 968;
  • 57) 0.704 976 118 670 163 968 × 2 = 1 + 0.409 952 237 340 327 936;
  • 58) 0.409 952 237 340 327 936 × 2 = 0 + 0.819 904 474 680 655 872;
  • 59) 0.819 904 474 680 655 872 × 2 = 1 + 0.639 808 949 361 311 744;
  • 60) 0.639 808 949 361 311 744 × 2 = 1 + 0.279 617 898 722 623 488;
  • 61) 0.279 617 898 722 623 488 × 2 = 0 + 0.559 235 797 445 246 976;
  • 62) 0.559 235 797 445 246 976 × 2 = 1 + 0.118 471 594 890 493 952;
  • 63) 0.118 471 594 890 493 952 × 2 = 0 + 0.236 943 189 780 987 904;
  • 64) 0.236 943 189 780 987 904 × 2 = 0 + 0.473 886 379 561 975 808;
  • 65) 0.473 886 379 561 975 808 × 2 = 0 + 0.947 772 759 123 951 616;
  • 66) 0.947 772 759 123 951 616 × 2 = 1 + 0.895 545 518 247 903 232;
  • 67) 0.895 545 518 247 903 232 × 2 = 1 + 0.791 091 036 495 806 464;
  • 68) 0.791 091 036 495 806 464 × 2 = 1 + 0.582 182 072 991 612 928;
  • 69) 0.582 182 072 991 612 928 × 2 = 1 + 0.164 364 145 983 225 856;
  • 70) 0.164 364 145 983 225 856 × 2 = 0 + 0.328 728 291 966 451 712;
  • 71) 0.328 728 291 966 451 712 × 2 = 0 + 0.657 456 583 932 903 424;
  • 72) 0.657 456 583 932 903 424 × 2 = 1 + 0.314 913 167 865 806 848;
  • 73) 0.314 913 167 865 806 848 × 2 = 0 + 0.629 826 335 731 613 696;
  • 74) 0.629 826 335 731 613 696 × 2 = 1 + 0.259 652 671 463 227 392;
  • 75) 0.259 652 671 463 227 392 × 2 = 0 + 0.519 305 342 926 454 784;
  • 76) 0.519 305 342 926 454 784 × 2 = 1 + 0.038 610 685 852 909 568;
  • 77) 0.038 610 685 852 909 568 × 2 = 0 + 0.077 221 371 705 819 136;
  • 78) 0.077 221 371 705 819 136 × 2 = 0 + 0.154 442 743 411 638 272;
  • 79) 0.154 442 743 411 638 272 × 2 = 0 + 0.308 885 486 823 276 544;
  • 80) 0.308 885 486 823 276 544 × 2 = 0 + 0.617 770 973 646 553 088;
  • 81) 0.617 770 973 646 553 088 × 2 = 1 + 0.235 541 947 293 106 176;
  • 82) 0.235 541 947 293 106 176 × 2 = 0 + 0.471 083 894 586 212 352;
  • 83) 0.471 083 894 586 212 352 × 2 = 0 + 0.942 167 789 172 424 704;
  • 84) 0.942 167 789 172 424 704 × 2 = 1 + 0.884 335 578 344 849 408;
  • 85) 0.884 335 578 344 849 408 × 2 = 1 + 0.768 671 156 689 698 816;
  • 86) 0.768 671 156 689 698 816 × 2 = 1 + 0.537 342 313 379 397 632;
  • 87) 0.537 342 313 379 397 632 × 2 = 1 + 0.074 684 626 758 795 264;
  • 88) 0.074 684 626 758 795 264 × 2 = 0 + 0.149 369 253 517 590 528;
  • 89) 0.149 369 253 517 590 528 × 2 = 0 + 0.298 738 507 035 181 056;
  • 90) 0.298 738 507 035 181 056 × 2 = 0 + 0.597 477 014 070 362 112;
  • 91) 0.597 477 014 070 362 112 × 2 = 1 + 0.194 954 028 140 724 224;
  • 92) 0.194 954 028 140 724 224 × 2 = 0 + 0.389 908 056 281 448 448;
  • 93) 0.389 908 056 281 448 448 × 2 = 0 + 0.779 816 112 562 896 896;
  • 94) 0.779 816 112 562 896 896 × 2 = 1 + 0.559 632 225 125 793 792;
  • 95) 0.559 632 225 125 793 792 × 2 = 1 + 0.119 264 450 251 587 584;

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:


0.000 000 000 000 176 563(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 1011 0100 0111 1001 0101 0000 1001 1110 0010 011(2)

6. Positive number before normalization:

0.000 000 000 000 176 563(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 1011 0100 0111 1001 0101 0000 1001 1110 0010 011(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:


0.000 000 000 000 176 563(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 1011 0100 0111 1001 0101 0000 1001 1110 0010 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 1011 0100 0111 1001 0101 0000 1001 1110 0010 011(2) × 20 =

1.1000 1101 1001 0101 1010 0011 1100 1010 1000 0100 1111 0001 0011(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 1001 0101 1010 0011 1100 1010 1000 0100 1111 0001 0011


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 1001 0101 1010 0011 1100 1010 1000 0100 1111 0001 0011 =

1000 1101 1001 0101 1010 0011 1100 1010 1000 0100 1111 0001 0011


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 1001 0101 1010 0011 1100 1010 1000 0100 1111 0001 0011


Decimal number -0.000 000 000 000 176 563 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0101 1010 0011 1100 1010 1000 0100 1111 0001 0011

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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