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.
#589_C

#import time

#start = time.time()

import math

md = (10 ** 9) + 7

def bin(n):
    l = math.floor(math.log(n, 2))
    s = ""
    while l >= 0:
        if n >= 2 ** l:
            n -= 2 ** l
            s += "1"
        else:
            s += "0"
        l -= 1
    return s

def ex(n, m):
    bn = bin(m)
    sm = n
    sms = []
    ssm = 1
    for i in range(len(bn) - 1, -1, -1):
        if bn[i] == "1":
            ssm = (ssm * sm) % md
        sm = (sm * sm) % md
    return ssm

ln = [int(i) for i in input().split(" ")]
x = ln[0]
n = ln[1]

div = []

def pf(n):
    for i in range(2, int(n ** 0.5) + 1):
        if n % i == 0:
            return [i] + pf(n // i)
    return [n]

dv = pf(x)

dv = sorted(dv)

ndv = [[dv[0], 1]]

for i in range(1, len(dv)):
    if dv[i] != dv[i - 1]:
        ndv.append([dv[i], 1])
    else:
        ndv[-1][1] += 1

sm = 1
for i in range(0, len(ndv)):
    nm = math.floor(math.log(n, ndv[i][0])) + 1
    for j in range(1, nm + 1):
        fs = n // (ndv[i][0] ** j)
        if fs == 0:
            continue
        nm = ndv[i][0]

        nn = ex(nm, fs)

        sm = (sm * nn) % md
        #print(sm, i)
print(sm)
