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 <bits/stdc++.h>
using namespace std;
using ll = long long;
using P = pair<ll, ll>;

const int mod = 1000000007;
struct mint {
    ll x;  // typedef long long ll;
    mint(ll x = 0) : x((x % mod + mod) % mod) {}
    mint operator-() const { return mint(-x); }
    mint& operator+=(const mint a) {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator-=(const mint a) {
        if ((x += mod - a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator*=(const mint a) {
        (x *= a.x) %= mod;
        return *this;
    }
    mint operator+(const mint a) const { return mint(*this) += a; }
    mint operator-(const mint a) const { return mint(*this) -= a; }
    mint operator*(const mint a) const { return mint(*this) *= a; }
    mint pow(ll t) const {
        if (!t) return 1;
        mint a = pow(t >> 1);
        a *= a;
        if (t & 1) a *= *this;
        return a;
    }

    // for prime mod
    mint inv() const { return pow(mod - 2); }
    mint& operator/=(const mint a) { return *this *= a.inv(); }
    mint operator/(const mint a) const { return mint(*this) /= a; }
};
istream& operator>>(istream& is, mint& a) { return is >> a.x; }
ostream& operator<<(ostream& os, const mint& a) { return os << a.x; }

// combination mod prime
// https://www.youtube.com/watch?v=8uowVvQ_-Mo&feature=youtu.be&t=1619
struct combination {
    vector<mint> fact, ifact;
    combination(int n) : fact(n + 1), ifact(n + 1) {
        assert(n < mod);
        fact[0] = 1;
        for (int i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i;
        ifact[n] = fact[n].inv();
        for (int i = n; i >= 1; --i) ifact[i - 1] = ifact[i] * i;
    }
    mint operator()(int n, int k) {
        if (k < 0 || k > n) return 0;
        return fact[n] * ifact[k] * ifact[n - k];
    }
};

void solve(long long N, long long M) {
    ll num = 2;
    vector<ll> factors;
    ll sum = 0;
    while (M >= num * num) {
        ll temp = 0;
        while (M % num == 0) {
            M /= num;
            temp++;
        }
        if (temp) {
            // cerr << "M=" << M << "num=" << num << "temp=" << temp << endl;
            factors.push_back(temp);
            sum += temp;
        }
        num++;
    }
    if (M > 1) {
        factors.push_back(1);
        sum++;
    }

    combination comb = combination(int(1e6) + 5);
    mint ans = 1;

    for (ll fact : factors) {
        ans *= comb(fact + N - 1, fact);
    }

    cout << ans << endl;
}

// Generated by 1.1.7.1 https://github.com/kyuridenamida/atcoder-tools  (tips: You use the default template now. You can remove this line by using your custom template)
int main() {
    long long N;
    scanf("%lld", &N);
    long long M;
    scanf("%lld", &M);
    solve(N, M);
    return 0;
}