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

#include <algorithm>
#include <bitset>
#include <cmath>
#include <cstring>
#include <deque>
#include <functional>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stdlib.h>
#include <string>
#include <utility>
#include <vector>
#define rep(i, n) for (int i = 0; i < (n); i++)
#define loop(i, x, n) for (int i = (x); i < (n); i++)
#define all(v) (v).begin(), (v).end()
#define int long long
using namespace std;
const int MOD = 1e9 + 7;
const int INF = 1e18;
template<typename T> void cmax(T &a, T b) { a = max(a, b); }
template<typename T> void cmin(T &a, T b) { a = min(a, b); }

int modinv(int a, int m) {
  int b = m, u = 1, v = 0;
  while (b) {
    int t = a / b;
    a -= t * b;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  u %= m;
  if (u < 0) u += m;
  return u;
}

map<int, int> prime_fac(int n) {
  map<int, int> ret;
  for (int i = 2; i * i <= n; i++) {
    while (n % i == 0) {
      ret[i]++;
      n /= i;
    }
  }
  if (n != 1) ret[n] = 1;
  return ret;
}
int fact[200005];
int inv[200005];

signed main() {
  int n, m;
  cin >> n >> m;
  map<int, int> mp = prime_fac(m);
  fact[0] = 1;
  inv[0] = 1;
  loop(i, 1, 200005) {
    fact[i] = fact[i - 1] * i;
    fact[i] %= MOD;
    inv[i] = modinv(fact[i], MOD);
  }

  int ans = 1;
  for (auto x : mp) {
    int t = x.second;
    // cout << x.first << ' ' << t << endl;
    ans *= fact[t + n - 1] * inv[t] % MOD * inv[n - 1] % MOD;
  }
  cout << ans % MOD << endl;
  return 0;
}