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 simple(d):
    if pow(2, d - 1, d) == 1:
        return [d]
    p = []
    sd = d
    i = 2
    while d != 1 or i < d // 2:
        if d % i == 0:
            p.append(i)
            d /= i
            while d % i == 0:
                d /= i
        if pow(2, int(d) - 1, int(d)) == 1:
            p.append(int(d))
            return p
        i += 1
    if len(p) == 0:
        p.append(sd)
    return p
    
x, g = tuple(map(int, input().split(" ")))
r = 1
r_n = 1
c = 1
power = 0
p = simple(x)
for j in p:
    i = j
    power = 0
    while g // i > 0: 
       #r *= j ** (g // i)
       power += (g // i)
       #print(r , i, g//i)
       i *= j
       c += 1
    #print(j, power)
    r_n *= pow(j, power, 1000000007)
    
    
print(r_n % 1000000007)