Problem:
Score : 400 points

Problem Statement

Given are positive integers A and B.

Let us choose some number of positive common divisors of A and B.

Here, any two of the chosen divisors must be coprime.

At most, how many divisors can we choose?

Definition of common divisor

An integer d is said to be a common divisor of integers x and y when d divides both x and y.

Definition of being coprime

Integers x and y are said to be coprime when x and y have no positive common divisors other than 1.

Definition of dividing

An integer x is said to divide another integer y when there exists an integer \alpha such that y = \alpha x.

Constraints

All values in input are integers.

1 \leq A, B \leq 10^{12}

Input

Input is given from Standard Input in the following format:

A B

Output

Print the maximum number of divisors that can be chosen to satisfy the condition.

Sample Input 1

12 18

Sample Output 1

3

12 and 18 have the following positive common divisors: 1, 2, 3, and 6.

1 and 2 are coprime, 2 and 3 are coprime, and 3 and 1 are coprime, so we can choose 1, 2, and 3, which achieve the maximum result.

Sample Input 2

420 660

Sample Output 2

4

Sample Input 3

1 2019

Sample Output 3

1

1 and 2019 have no positive common divisors other than 1.

Solution in C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.IO;
using System.Text;
using static System.Console;
using static System.Math;

namespace AtCodeeeer
{
    class AtCOOOOOOOOOOOder
    {
        static void Main(string[] args)
        {
            long[] s = ReadLine().Split(' ').Select(long.Parse).ToArray();
            long a = s[0];
            long b = s[1];
            long gcd = Gcd(a, b);
            if(gcd == 1)
            {
                WriteLine(1);
                return;
            }
            var ansList = new List<int>();
            var ls = new List<int>();
            ls.Add(1);
            int[] primes = { 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,
 61,67,71,73,79,83,89,97,101,103,107,109,113,127,
 131,137,139,149,151,157,163,167,173,179,181,191,
 193,197,199,211,223,227,229,233,239,241,251,257,
 263,269,271,
277,
281,
283,
293,
307,
311,
313,
317,
331,
337,
347,
349,
353,
359,
367,
373,
379,
383,
389,
397,
401,
409,
419,
421,
431,
433,
439,
443,
449,
457,
461,
463,
467,
479,
487,
491,
499,
503,
509,
521,
523,
541,
547,
557,
563,
569,
571,
577,
587,
593,
599,
601,
607,
613,
617,
619,
631,
641,
643,
647,
653,
659,
661,
673,
677,
683,
691,
701,
709,
719,
727,
733,
739,
743,
751,
757,
761,
769,
773,
787,
797,
809,
811,
821,
823,
827,
829,
839,
853,
857,
859,
863,
877,
881,
883,
887,
907,
911,
919,
929,
937,
941,
947,
953,
967,
971,
977,
983,
991,
997,
1009,
1013,
1019,
1021,
1031,
1033,
1039,
1049,
1051,
1061,
1063,
1069,
1087,
1091,
1093,
1097,
1103,
1109,
1117,
1123,
1129,
1151,
1153,
1163,
1171,
1181,
1187,
1193,
1201,
1213,
1217,
1223,
1229,
1231,
1237,
1249,
1259,
1277,
1279,
1283,
1289,
1291,
1297,
1301,
1303,
1307,
1319,
1321,
1327,
1361,
1367,
1373,
1381,
1399,
1409,
1423,
1427,
1429,
1433,
1439,
1447,
1451,
1453,
1459,
1471,
1481,
1483,
1487,
1489,
1493,
1499,
1511,
1523,
1531,
1543,
1549,
1553,
1559,
1567,
1571,
1579,
1583,
1597,
1601,
1607,
1609,
1613,
1619,
1621,
1627,
1637,
1657,
1663,
1667,
1669};
            ls.AddRange(primes);
            for (int i = 0; i <ls.Count; i++)
            {
                var target = ls[i];
                bool find = false;
                //targetが公約数である
                if (a % target == 0 && b % target == 0)
                {
                    //iがすべての他の要素と互いに素である
                    for (int j = 0; j < ansList.Count; j++)
                    {
                        if (Gcd(target,ansList[j]) != 1) find = true;
                    }
                    if(find == false)ansList.Add(target);
                }
            }
            WriteLine(ansList.Count);
        }

        /*
         * Gcd (Gross Common Dividing) 最大公約数
         * 
        */
        static long Gcd(long a, long b)
        {
            var v = new[] { a, b };
            while (v[1] != 0) { v = new[] { v[1], v[0] % v[1] }; }
            return v[0];
        }
    }
}