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: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
void fact(long a[], long c[], long d[], long m, long n, int c1, int d1);
void merge(long a1[], long a2[], long c[], long d[], long m, int d1);


int main(void)
{
	int j=0, k=0, c1=0, d1=0;
	long n, m, *c, *d, *a, ans=0;
	scanf("%ld%ld", &n, &m);
	for (long i = 1; i <= m; ++i)
	{
		if (m%i == 0)
		{
			c1++;
		}
	}
	c=(long *)malloc(sizeof(long)*c1);
	a=(long *)malloc(sizeof(long)*(int)(m+1));
	for (int i = 1; i <= m; ++i)
	{
		if (m%i == 0)
		{
			c[j]=i;
			j++;
		}
	}
	for (int i1 = 0; i1 < c1; i1++)
	{
		for (int i2 = 0; i2 < c1; i2++)
		{
			if ((m+c[i1]*c[i2])%(c[i1]*c[i2])==0)
			{
				d1++;
			}
		}
	}
	d=(long *)malloc(sizeof(long)*d1);
	for (int i1 = 0; i1 < c1; i1++)
	{
		for (int i2 = 0; i2 < c1; i2++)
		{
			if ((m+c[i1]*c[i2])%(c[i1]*c[i2])==0)
			{
				d[k]=10*i1+i2;
				k++;
			}
		}
	}
	printf("\n");
	fact(a, c, d, m, n, c1, d1);
	ans = a[m];
	printf("%ld\n", ans);
	free(c);
	free(d);
	free(a);
	return 0;
}

void fact(long a[], long c[], long d[], long m, long n, int c1, int d1)
{
	long *a1, *a2;
	if (n==1)
	{
		for (int i = 0; i < c1; ++i)
		{
			a[c[i]]=1;
		}
	}
	else
	{
		a1=(long *)malloc(sizeof(long)*(int)(m+1));
		a2=(long *)malloc(sizeof(long)*(int)(m+1));
		fact(a1, c, d, m, n/2, c1, d1);
		fact(a2, c, d, m, n-n/2, c1, d1);
		merge(a1, a2, c, d, m, d1);
		for (int i = 0; i < c1; ++i)
		{
			a[c[i]]=a1[c[i]];
		}
		free(a1);
		free(a2);
	}
	return;
}

void merge(long a1[], long a2[], long c[], long d[], long m, int d1)
{
	int s, t;
	long *a3, mod=1000000007;
	a3=(long *)malloc(sizeof(long)*(int)(m+1));
	for (int i = 0; i <= m; ++i)
	{
		a3[i] = 0;
	}
	for (int i = 0; i < d1; ++i)
	{
		a3[c[d[i]/10]*c[(d[i]+10)%10]] += a1[c[d[i]/10]]*a2[c[(d[i]+10)%10]];
		a3[c[d[i]/10]*c[(d[i]+10)%10]] = (a3[c[d[i]/10]*c[(d[i]+10)%10]]+mod)%mod;
	}
	for (int i = 0; i <= m; ++i)
	{
		a1[i]=a3[i];
	}
	free(a3);
	return;
}