Let's introduce some definitions that will be needed later.

Let prime(x) be the set of prime divisors of x. For example, prime(140) = \{ 2, 5, 7 \}, prime(169) = \{ 13 \}.

Let g(x, p) be the maximum possible integer p^k where k is an integer such that x is divisible by p^k. For example:

  * g(45, 3) = 9 (45 is divisible by 3^2=9 but not divisible by 3^3=27), 
  * g(63, 7) = 7 (63 is divisible by 7^1=7 but not divisible by 7^2=49). 



Let f(x, y) be the product of g(y, p) for all p in prime(x). For example:

  * f(30, 70) = g(70, 2) ⋅ g(70, 3) ⋅ g(70, 5) = 2^1 ⋅ 3^0 ⋅ 5^1 = 10, 
  * f(525, 63) = g(63, 3) ⋅ g(63, 5) ⋅ g(63, 7) = 3^2 ⋅ 5^0 ⋅ 7^1 = 63. 



You have integers x and n. Calculate f(x, 1) ⋅ f(x, 2) ⋅ … ⋅ f(x, n) mod{(10^{9} + 7)}.

Input

The only line contains integers x and n (2 ≤ x ≤ 10^{9}, 1 ≤ n ≤ 10^{18}) — the numbers used in formula.

Output

Print the answer.

Examples

Input


10 2


Output


2


Input


20190929 1605


Output


363165664


Input


947 987654321987654321


Output


593574252

Note

In the first example, f(10, 1) = g(1, 2) ⋅ g(1, 5) = 1, f(10, 2) = g(2, 2) ⋅ g(2, 5) = 2.

In the second example, actual value of formula is approximately 1.597 ⋅ 10^{171}. Make sure you print the answer modulo (10^{9} + 7).

In the third example, be careful about overflow issue.
# -*- coding: utf-8 -*-
# if x < y:
#     z = x
# elif True:
#     z = y
# else:
#     z = 100
import sys,os,math
from collections import Counter, defaultdict
import bisect
from sys import stdin, stdout
from itertools import repeat


# n, k = map(int, raw_input().split())
# da = map(int, raw_input().split())
# db = map(int, raw_input().split())

def main():
    ans = 1
    x, n = map(int, raw_input().split())
    fac = set()
    xx = x
    for i in range(2, int(math.sqrt(x)+1)):
        if xx%i==0:
            fac.add(i)
            while xx%i == 0:
                xx/=i
    # print fac
    if xx!=1:
        fac.add(xx)

    mod = 1000000007
    def getmod(a, b, m=1000000007):
        res = 1
        while b:
            if b&1:
                res  = (res*a)%m
            a = (a*a)%m
            b >>= 1
        return res
    for i in fac:
        tp = i
        while tp <= n:
            ti = n/tp
            ti -= ti/i
            ans *= getmod(tp, ti)
            ans %= mod
            tp *= i
    print ans%mod
main()
