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.
"""
#If FastIO not needed, used this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""

import os
import sys
from io import BytesIO, IOBase
import heapq as h 
from bisect import bisect_left, bisect_right

from types import GeneratorType
BUFSIZE = 8192
class FastIO(IOBase):
    newlines = 0
 
    def __init__(self, file):
        import os
        self.os = os
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
 
    def read(self):
        while True:
            b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
 
    def readline(self):
        while self.newlines == 0:
            b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
 
    def flush(self):
        if self.writable:
            self.os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)
 
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

import collections as col
import math, string

def getInts():
    return [int(s) for s in input().split()]

def getInt():
    return int(input())

def getStrs():
    return [s for s in input().split()]

def getStr():
    return input()

def listStr():
    return list(input())

MOD = 10**9+7

def prime_factors(n):
    factors = set()
    num = 0
    while n % 2 == 0:
        num += 1
        n //= 2
    if num:
        factors.add(2)
    i = 3
    while n > 1 and i*i <= n:
        num = 0
        while n % i == 0:
            num += 1
            n //= i
        if num:
            factors.add(i)
        i += 2
    if n > 2:
        factors.add(n)
    return factors


"""
prime(x) = set of primes that divide x
g(x,p) is the highest power of p that divides x
f(x,y) is g(p1,y)*g(p2,y)*...*g(pn,y) for all pi in prime(x)
we want product f(x,1)...f(x,n) mod 10**9+7

g(p1,1)*g(p2,1)*...*g(pn,1) * g(p1,2)*g(p2,2)*...*g(pn,n) * ... 

Up to and including N, how many powers of each factor do we see?
1 for every multiple
1 more for every multiple of its square
1 more for every multiple of its cube
etc
"""

#2**0 * 5**0 * 2**1 * 5**0 * 2**0 * 5**0 * 2**2 * 5**0 * 2**0 * 5**1
#= 2**3 * 5**1
#= 40

def solve():
    X, N = getInts()
    primex = prime_factors(X)
    ans = 1
    for p in primex:
        p_pow = p
        while p_pow <= N:
            ans *= pow(p,N//p_pow,MOD)
            ans %= MOD
            p_pow *= p
    return ans
    
#for _ in range(getInt()):
print(solve())