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 Python (3.4.3) (judge verdict: Runtime Error)

import sys
stdin = sys.stdin

sys.setrecursionlimit(10**5)

def li(): return map(int, stdin.readline().split())
def li_(): return map(lambda x: int(x)-1, stdin.readline().split())
def lf(): return map(float, stdin.readline().split())
def ls(): return stdin.readline().split()
def ns(): return stdin.readline().rstrip()
def lc(): return list(ns())
def ni(): return int(stdin.readline())
def nf(): return float(stdin.readline())

from collections import Counter

def factorize(n: int):
    d = Counter()
    m= 2
    
    while m*m <= n:
        while n%m == 0:
            n //= m
            d[m] += 1
            
        m += 1
        
    if n > 1:
        d[n] += 1
        
    return d

# nの逆元のリスト
def inv_mod(n:int, mod:int) -> list:
    inv = [0,1]
    for i in range(2,n+1):
        inv.append(mod - ((mod//i)*inv[mod%i]) % mod)
    return inv
       
# nの階乗のリスト
def fact(n:int, mod:int) -> list:
    fac = [1,1]
    res = 1
    for i in range(2,n+1):
        res = res*i%mod
        fac.append(res)
    return fac

# nの階乗の逆元のリスト
def fact_inv(n:int, inv:list, mod:int) -> list:
    facInv = [1,1]
    for i in range(2,n+1):
        facInv.append(facInv[i-1]*inv[i] % mod)
    return facInv
        
# 二項係数
def nCr(n:int, r:int, mod:int, fac:list, facInv:list) -> int:
    if not (0<=r and r<=n):
        return 0
    
    return ((fac[n]*facInv[r]) % mod) * facInv[n-r] % mod

# 重複組み合わせ
def nHr(n:int, r:int, mod:int, fac:list, facInv:list) -> int:
    if r<0 or n<0:
        return 0
    
    else:
        return nCr(n+r-1,r,mod,fac,facInv)

n,m = li()
MOD = 10**9+7

inv = inv_mod(10**5, MOD)
fac = fact(10**5, MOD)
facinv = fact_inv(10**5,inv,MOD)

md = factorize(m)

ans = 1
for k,v in md.items():
    ans *= nHr(n,v,MOD,fac,facinv)
    ans %= MOD
    
print(ans)