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.
def isprime(n):
	if(n == 1):
		return False
	if(n == 2):
		return True
	i = 2
	while(i*i <= n):
		if(n % i == 0):
			return False
		i += 1
	return True
def primefactors(n):
	s = set()
	if(isprime(n)):
		s.add(n)
		return s
	i = 2
	while(i*i <= n):
		if(n % i == 0 and isprime(i)):
			s.add(i)
			while(n % i == 0):
				n //= i
			if(isprime(n)):
				s.add(n)
		i += 1
	return s
x , n = map(int , input().split(' '))
s = primefactors(x)                                       
def maxp(n , i):
	ans = 0
	t = i
	while(n//i > 0):
		ans += n // i;
		i *= t
	return ans
def power(x, y, p) : 
	res = 1
	x = x % p  
	if (x == 0) : 
		return 0
	while (y > 0) : 
		if ((y & 1) == 1) : 
			res = (res * x) % p 
		y = y >> 1 
		x = (x * x) % p 
	return res 
	
ans = 1
for i in s:
	ans = (ans * power(i , maxp(n , i) , 1000000007)) % 1000000007
print(ans)