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.
from sys import stdin, stdout
 
ii = lambda: int(stdin.readline())
mi = lambda: map(int, stdin.readline().strip().split())
li = lambda: list(mi())

MOD = int(1000000007)

def binpow(a, b, m):
	a %= m
	res = 1
	while (b > 0):
		if (b & 1):
			res = res * a % m
		a = a * a % m
		b >>= 1
	return res

x, n = mi()
fac = []
tmp = int(x)
if (tmp % 2 == 0):
	fac.append(2)
	while(tmp % 2 == 0):
		tmp = tmp // 2
i = 3
while (i * i <= tmp):
	if (tmp % i == 0):
		fac.append(i)
		while(tmp % i == 0):
			tmp = tmp // i
	i += 2

if (tmp > 2):
	fac.append(tmp)
ans = 1
for p in fac:
	prod = p
	i = 1
	fin = 0
	while (prod <= n):
		times = (n // prod) - (n // (prod * p))
		fin += (i * times)
		prod *= p
		i += 1
	ans = (ans * binpow(p, fin, MOD)) % MOD
print(ans)
