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 Rust (1.15.1) (judge verdict: Accepted)

#[allow(unused_macros)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };

    ($next:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, bytes) => {
        read_value!($next, String).into_bytes()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

#[allow(unused_imports)]
use std::cmp::{max, min};
#[allow(unused_imports)]
use std::collections::{BinaryHeap, HashMap};

#[allow(dead_code)]
type Graph = Vec<Vec<usize>>;

const M: usize = 1000000007;

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

fn gcd(a: usize, b: usize) -> usize {
    if b != 0 {
        gcd(b, a % b)
    } else {
        a
    }
}

fn lcm(a: usize, b: usize) -> usize {
    a / gcd(a, b) * b
}

pub fn extgcd(a: i64, b: i64) -> (i64, i64, i64) {
    if b == 0 {
        (a, 1, 0)
    } else {
        let (gcd, x, y) = extgcd(b, a % b);
        (gcd, y, x - (a / b) * y)
    }
}

pub fn mod_inverse(a: usize, m: usize) -> usize {
    let (_, x, _) = extgcd(a as i64, m as i64);
    ((m as i64 + x) as usize % m) % m
}

fn main() {
    input! {
        n: usize, m: usize,
    }

    let mut hash = HashMap::new();

    let mut mm = m;
    let mut p = 2;
    while p * p <= m {
        while mm % p == 0 {
            *hash.entry(p).or_insert(0) += 1;
            mm /= p;
        }
        p += 1;
    }
    if mm > 1 {
        hash.entry(mm).or_insert(1);
    }

    let mut ans = 1;

    // let mut comb = vec![vec![0; 100100]; 101000];
    // for i in 0..1000 {
    //     for j in 0..i+1{
    //         if j == 0 || j == i {
    //             comb[i][j] = 1;
    //         } else {
    //             comb[i][j] = (comb[i-1][j-1] + comb[i-1][j]) % M;
    //         }
    //     }
    // }

    for (_, v) in &hash {
        // ans *= comb[*v + n - 1][*v];
        let mut c = 1;
        for i in 1..v+1 {
            c *= v + n - i;
            c %= M;
            c *= mod_inverse(i, M);
            c %= M;
        }
        ans *= c;
        ans %= M;
    }


    println!("{}", ans);
}