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 (Clang 3.8.0) (judge verdict: Time Limit Exceeded)


#include <stdio.h>

int main(void){

	long A,B;
	
	long i,c;
	
	int ret;
	
	ret = scanf("%ld %ld",&A,&B);
	
	i=1;
	c=1;
	
	int fA,fB;

	while( A > i && B > i ){
		
		i++;
			
		fA=0;
		fB=0;

		while ((A % i) == 0){
			A = A / i;
			fA=1;
		}
		
		while ((B % i) == 0){
			B = B / i;
			fB=1;
		}
		
		if (fA==1 && fB==1){
			
			c++;
		}
		
	}
	
	printf("%ld\n",c);
	

}