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.
[x, n] = input().split(' ')
x = int(x)
n = int(n)
mod = 1000000007
primes = []

def is_prime(n):
	if(n<2): return False
	if(n==2): return True
	if(n%2==0): return False
	i = 3
	while(i*i<=n):
		if(n%i==0): return False
		i += 2
	return True

def fast_pow(a, b):
	r = 1
	while(b>0):
		if(b&1==1): r = (r*a)%mod
		a = (a*a)%mod
		b //= 2
	return r

def query(p):
	ans = 1
	v = p
	while(v<=n):
		ans = (ans*fast_pow(p, n//v))%mod
		v *= p
	return ans

ans = 1
if(is_prime(x)):
	ans = query(x)
else:
	i = 2
	while(x>1 and i*i<=x):
		if(x%i!=0):
			i += 1
			continue
		ans = (ans*query(i))%mod
		while(x%i==0): x //= i
		i += 1
	if(x>1): ans = (ans*query(x))%mod
print(ans)