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.
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
import math
x,n=list(map(int,input().split()))
z=(10**9)+7
def primeFactorization(a):
    dict1={}
    lis=[]
    while a%2==0:
        if 2 not in dict1.keys():
            dict1[2]=1
            lis.append(2)
        a=a//2
    b=int(math.sqrt(a))
    for i in range(3,b+1,2):
        while a%i==0:
            if i not in dict1.keys():
                dict1[i]=1
                lis.append(i)
            a=a//i
    if a>2 and (a not in dict1.keys()):
        dict1[a]=1
        lis.append(a)
    return lis
def countOccurence(r,p):
    val=p
    ans=0
    while True:
        a=r//val
        val*=p
        if a>0:
            ans+=a
        else:
            break
    return ans
def power(p,q,z):
    if q==0:
        return 1
    temp=power(p,q//2,z)
    if q%2==0:
        return (temp*temp)%z
    return (temp*temp*p)%z
arr=primeFactorization(x)
l=len(arr)
ans=1
for i in range(l):
    num=countOccurence(n,arr[i])
    temp=power(arr[i],num,z)
    ans=(ans*temp)%z
print(ans)