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.
# MOD = 10**180
MOD = 1000000007

x, n = map(int, input().split())

def po(n, p):
	if p == 0: return 1
	if p == 1: return n % MOD
	p2 = p//2
	if p % 2 == 0: return (po(n, p2)**2) % MOD
	else: return (n * po(n, p2)**2) % MOD

def f(s, x, n):
	if x == 1: return 1
	sx = int(x**0.5)
	for p in range(s, sx+1):
		if x % p == 0:
			while x % p == 0:
				x //= p
			res = f(p+1, x, n)
			m = 0
			pp = p
			while n // pp > 0:
				m = (m + n // pp)
				pp *= p
			pm = po(p, m)
			res = (res*pm) % MOD
			return res
	m = 0
	pp = x
	while n // pp > 0:
		m = (m + n // pp)
		pp *= x
	return po(x, m) % MOD

res = f(2, x, n)
print(res)