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.
import sys
import os
import math

def READ(fileName):
    sys.stdin = open('in.txt')


def sieve():
    primes = []
    M = 100007
    marked = [0 for x in range(M)]
    sz = int(math.sqrt(M))
    for i in range(3, sz, 2):
        if marked[i] == 0:
            for j in range(i*i, M, 2*i):
                marked[j] = 1

    primes.append(2)

    for i in range(3, M, 2):
        if marked[i] == 0:
            primes.append(i)

    return primes 

def g(x, p):
    cnt = 0

    while x != 0:
        cnt += x//p
        x //= p

    res = pow(p, cnt, 1000000007)
    return res

if __name__ == "__main__":
    #READ('in.txt')
    mod = 1000000007
    while True:
        try:
            x, n = map(int, input().split())
        except EOFError:
            break
        
        primes = sieve()
        factors = []

        for prime in primes:
            if x % prime == 0:
                factors.append(prime)
                while x % prime == 0:
                    x //= prime

        if x > 1:
            factors.append(x)
        #print(factors)
        res = 1
        for factor in factors:
            #res *= g(n, factor)
            res = ((res % mod) * (g(n, factor) % mod)) % mod
        
        print(res)

    pass