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.
'''input
20190929 1605

'''
import sys
from collections import defaultdict as dd
from itertools import  permutations as pp
from itertools import combinations as cc
from collections import Counter as ccd
from random import randint as rd
from bisect import bisect_left as bl
from  heapq import heappush as hpush
from heapq import heappop as hpop
mod=10**9+7

def ri(flag=0):
	if flag==0:
		return [int(i) for i in sys.stdin.readline().split()]
	else:
		return int(sys.stdin.readline())


fac = set()

n , m = ri()

temp = n
for i in range(2,10**6):
	while temp%i==0:
		fac.add(i)
		temp = temp//i

if (temp!=1):
	fac.add(temp)


ans = 1

for x in fac:
	temp = x
	while(temp <=m):
		count = m//temp
		ans = (ans * pow(x, count,mod))%mod;
		temp = temp*x

print(ans%mod)