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.
from collections import defaultdict, deque, Counter
from sys import stdin, stdout
from heapq import heappush, heappop
import math
import io
import os
import math
import bisect

#?############################################################


def isPrime(x):
    for i in range(2, x):
        if i*i > x:
            break
        if (x % i == 0):
            return False
    return True

#?############################################################


def ncr(n, r, p):
    num = den = 1
    for i in range(r):
        num = (num * (n - i)) % p
        den = (den * (i + 1)) % p
    return (num * pow(den, p - 2, p)) % p


#?############################################################

def primeFactors(n):
    l = []
    while n % 2 == 0:
        l.append(2)
        n = n // 2
    for i in range(3, int(math.sqrt(n))+1, 2):
        while n % i == 0:
            l.append(i)
            n = n // i
    if n > 2:
        l.append(n)
    return list(set(l))


#?############################################################

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

#?############################################################


def sieve(n):
    prime = [True for i in range(n+1)]
    p = 2
    while (p * p <= n):
        if (prime[p] == True):
            for i in range(p * p, n+1, p):
                prime[i] = False
        p += 1
    return prime


#?############################################################

def digits(n):
    c = 0
    while (n > 0):
        n //= 10
        c += 1
    return c

#?############################################################


def ceil(n, x):
    if (n % x == 0):
        return n//x
    return n//x+1

#?############################################################


def mapin():
    return map(int, input().split())

#?############################################################


input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# python3 15.py<in>op
p = 1000000007
t = 1
for _ in range(t):
    x, n = mapin()
    a = primeFactors(x)
    ans = 1
    # print(a)
    

    for i in a:
        x = i

        while(x<=n):
            temp = n//x
            temp2 = power(i, temp, p)
            
            ans*=temp2
            ans%=p
            x *=i
    print(ans%p)
    