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.
ans = 1
MOD = 10 ** 9 + 7
def prime_factorize(num):
    """
    This function performs prime factorization on the input natural number.
    The result is returned in the form of a dictionary with the prime number as the key
    and its number as the value.
    :param num:
    :return prime_factor: Dictionary with the prime number as the key and its number as the value.
    """
    prime_factor = {}
    i = 2
    while i ** 2 <= num:
        while num % i == 0:
            num //= i
            if i in prime_factor.keys():
                prime_factor[i] += 1
            else:
                prime_factor[i] = 1
        i += 1
    if num > 1:
        prime_factor[num] = 1
    return prime_factor


def divisorize(fct):
    b, e = fct.pop()  # base, exponent
    pre_div = divisorize(fct) if fct else [[]]
    suf_div = [[(b, k)] for k in range(e + 1)]
    return [pre + suf for pre in pre_div for suf in suf_div]


def num(fct):
    a = 1
    for base, exponent in fct:
        a = a * base ** exponent
    return a

# prime_factor = prime_factorize(N)
# prime_factor = list(prime_factor.items())
# prime_list = []
# for div in divisorize(prime_factor):
#     prime_list.append(num(div))

x, n = map(int, input().split())
pf = prime_factorize(x)
tar = []
for k, v in pf.items():
    tar.append(k)

for i in range(len(tar)):
    t = tar[i]
    beki = []
    e = 1
    while t * e <= n:
        beki.append(t * e)
        e = t * e
    beki.sort(reverse=True)
    beki_cnt = 0
    for j in range(len(beki)):
        be = beki[j]
        ans *= pow(be , n // be - beki_cnt, MOD)
        ans %= MOD
        beki_cnt += n // be - beki_cnt
print(ans)

