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 (1.42.0) (judge verdict: Accepted)

use proconio::input;
use maplit::hashmap;

#[proconio::fastout]
fn main() {
    input! {
        N: usize,
        M: i128
    }
    const MOD: usize = 1000_000_000 + 7;

    let mut prime_divisors = hashmap!();
    let mut div = 2;
    let mut m = M;
    while div * div <= m {
        let count = prime_divisors.entry(div).or_insert(0);
        while m % div == 0 {
            *count += 1;
            m /= div;
        }
        div += 1;
    }
    if m != 1 {
        *(prime_divisors.entry(m).or_insert(0)) += 1;
    }

    let num_of_divisors: usize = prime_divisors.values().sum();
    let x = N - 1 + num_of_divisors;
    let mut inverse = vec![0, 1];
    let mut inverted_factorials = vec![1, 1];
    let mut factorials = vec![1, 1];
    for f in 2..=x {
        factorials.push(factorials.last().unwrap() * f % MOD);
        inverse.push(MOD - inverse[MOD % f] * (MOD / f) % MOD);
        inverted_factorials.push(inverted_factorials.last().unwrap() * inverse[f] % MOD);
    }

    let inverted_factorials: Vec<i128> = inverted_factorials.iter().map(|&x| x as i128).collect();
    let factorials: Vec<i128> = factorials.iter().map(|&x| x as i128).collect();
    let result = {
        let mut result = 1i128;
        for &x in prime_divisors.values() {
            result *= factorials[N - 1 + x];
            result %= MOD as i128;
            result *= inverted_factorials[N - 1];
            result %= MOD as i128;
            result *= inverted_factorials[x];
            result %= MOD as i128;
        }
        result
    };

    println!("{}", result)
}