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.
#!/usr/bin/env python3
import sys
from math import sqrt

def rint():
    return map(int, sys.stdin.readline().split())
#lines = stdin.readlines()

def find_prime_factors(num):
    d = set()
    i = 2
    while  i < int(sqrt(num)+1):
        while num%i == 0:
            d.add(i)
            num = num//i
        i += 1
    if num > 1:
        d.add(num)
    return d


def power_modulo(x, p, m):
    res = 1
    x = x%m
    while p > 0:
        if p%2:
            res = (res*x)%m

        p = p//2
        x = (x*x)%m
    return res


def solve():
    ans = 1
    for i in range(len(p)):
        pp = p[i]
        power = 0
        for j in range(10**20):
            power += n//pp
            if n//pp == 0:
                break
            pp *= p[i]
        ans = ans*power_modulo(p[i], power, 10**9+7)
    return ans%(10**9+7)


x, n = rint()

p = find_prime_factors(x)
p = list(p)
p.sort()
ans = solve()
print(ans)
