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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

const int N = 1000010;
typedef long long ll;

ll A, B;
int p[N], cnt, vis[N];

int main() {
	scanf("%lld%lld", &A, &B);
	ll ans = 1;
	for(int i = 2; i <= 1000000; ++i) {
		if(!vis[i]) {
			p[++cnt] = i;
		}
		for(int j = 1; j <= cnt && i * p[j] <= 1000000; ++j) {
			vis[i * p[j]] = 1;
			if(i % p[j] == 0) break;
		}
	}
	for(int i = 1; i <= cnt; ++i) {
		if(A % p[i] == 0 && B % p[i] == 0) {
			++ans;
			while(A % p[i] == 0) A /= p[i];
			while(B % p[i] == 0) B /= p[i];
		}
	}
	if(A > p[cnt] && B > p[cnt] && A == B) ++ans;
	printf("%d\n", ans);
}