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.
mod=int(1000000007);
def power(x,y,m):
    ans=1;
    while y>0:
        
        if y%2==1:
            ans=(ans*x)%m;
        
        y=y//2;
        x=(x*x)%m;
    
    return ans;

x,n=input().split();
x=int(x);n=int(n);
prime=[];

i=2;temp=x;
while i*i<=x:
    
    if temp%i==0:
        prime.append(i);
        while temp%i==0:
            temp=temp//i;
    i+=1
    
if temp!=1:
    prime.append(temp);
    
'''for it in prime:
    print(it,end=" ");
print();'''
ans=1;
for it in prime:
    i=1
    while (it**i)<=n:
        #print(it**i,end=" ");
        k=n//(it**i);
        ans=(ans*(power(it,k,mod)))%mod;
        i+=1
#print();
print(int(ans));