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 9.2.1) (judge verdict: Accepted)

#include <stdio.h>

#define Mod 1000000007

long long div_mod(long long x, long long y, long long z)
{
	if (x % y == 0) return x / y;
	else return (div_mod((1 + x / y) * y - x, (z % y), y) * z + x) / y;
}

int factorize(int n, int p[], int m[])
{
	int i, k;
	for (i = 2, k = 0; i * i <= n; i++) {
		if (n % i != 0) continue;
		p[k] = i;
		for (m[k] = 0; n % p[k] == 0; m[k]++, n /= p[k]);
		k++;
	}
	if (n > 1) {
		p[k] = n;
		m[k++] = 1;
	}
	return k;
}

int main()
{	
	int N, M;
	scanf("%d %d", &N, &M);
	if (N == 1 || M == 1) {
		printf("1\n");
		fflush(stdout);
		return 0;
	}

	int k, p[100], m[100];
	k = factorize(M, p, m);
	
	int h, i, j, l, n, bit[100];
	for (i = 1, n = 0, bit[0] = 1; i <= k; i++) {
		n += m[i-1];
		bit[i] = bit[i-1] * (m[i-1] + 1);
	}
	if (n > N) n = N;
	long long dp[50][2000] = {};
	for (i = 1, dp[0][0] = 1; i <= n; i++) {
		for (j = bit[k] - 1; j > 0; j--) {
			for (l = j - 1; l >= 0; l--) {
				for (h = 0; h < k; h++) if (j % bit[h+1] / bit[h] < l % bit[h+1] / bit[h]) break;
				if (h >= k) dp[i][j] += dp[i-1][l];
			}
			dp[i][j] %= Mod;
		}
	}
	
	long long ans = 0, fact[100001], fact_inv[100001];
	for (i = 1, fact[0] = 1; i <= N; i++) fact[i] = fact[i-1] * i % Mod;
	for (i = N - 1, fact_inv[N] = div_mod(1, fact[N], Mod); i >= 0; i--) fact_inv[i] = fact_inv[i+1] * (i + 1) % Mod;
	for (i = 1; i <= n; i++) ans = (ans + dp[i][bit[k]-1] * fact_inv[i] % Mod * fact_inv[N-i]) % Mod;
	printf("%lld\n", ans * fact[N] % Mod);
	fflush(stdout);
	return 0;
}