Problem:
Score : 400 points

Problem Statement

You are given positive integers N and M.

How many sequences a of length N consisting of positive integers satisfy a_1 \times a_2 \times ... \times a_N = M? Find the count modulo 10^9+7.

Here, two sequences a' and a'' are considered different when there exists some i such that a_i' \neq a_i''.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq M \leq 10^9

Input

Input is given from Standard Input in the following format:

N M

Output

Print the number of the sequences consisting of positive integers that satisfy the condition, modulo 10^9 + 7.

Sample Input 1

2 6

Sample Output 1

4

Four sequences satisfy the condition: \{a_1, a_2\} = \{1, 6\}, \{2, 3\}, \{3, 2\} and \{6, 1\}.

Sample Input 2

3 12

Sample Output 2

18

Sample Input 3

100000 1000000000

Sample Output 3

957870001

Solution in C (GCC 5.4.1) (judge verdict: Runtime Error)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
void fact(long a[], long c[], long m, long n, int c1);
void merge(long a1[], long a2[], long c[], long m, int c1);


int main(void)
{
	int k, c1=0;
	long n, m, *c, *a, ans=0;
	scanf("%ld%ld", &n, &m);
	c=(long *)malloc(sizeof(long)*(int)(2*(int)sqrt(m)));
	for (long i = 1; i <= (int)sqrt(m); ++i)
	{
		if (m%i == 0)
		{
			c[c1]=i;
			c1++;
		}
	}
	for (long i = (int)sqrt(m); i >= 1; i--)
	{
		if ((int)pow(i, 2) == m) i--;
		if (m%i == 0)
		{
			c[c1]=m/i;
			c1++;
		}
	}
	a=(long *)malloc(sizeof(long)*c1);
	fact(a, c, m, n, c1);
	ans = a[c1-1];
	printf("%ld\n", ans);
	free(c);
	free(a);
	return 0;
}


void fact(long a[], long c[], long m, long n, int c1)
{
	long a1[c1], a2[c1];
	if (n==1)
	{
		for (int i = 0; i < c1; ++i)
		{
			a[i]=1;
		}
	}
	else
	{
		fact(a1, c, m, n/2, c1);
		fact(a2, c, m, n-n/2, c1);
		merge(a1, a2, c, m, c1);
		for (int i = 0; i < c1; ++i)
		{
			a[i]=a1[i];
		}
	}
	return;
}

void merge(long a1[], long a2[], long c[], long m, int c1)
{
	int k;
	long a3[c1], mod=1000000007;
	for (int i = 0; i < c1; ++i)
	{
		a3[i] = 0;
	}
	for (int i = 0; i < c1/2+1; ++i)
	{
		for (int j = i; j < c1; ++j)
		{
			if (((m+c[i]*c[j])%(c[i]*c[j]) == 0) && (c[i]*c[j] <= m))
			{
				k=0;
				while (c[k] != c[i]*c[j]) k++;
				a3[k] += a1[i]*a2[j];
				if (i != j) a3[k] += a1[j]*a2[i];
				a3[k] = (a3[k]+mod)%mod;
			}
		}
	}
	for (int i = 0; i < c1; ++i)
	{
		a1[i]=a3[i];
	}
	return;
}