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 java.util.*; 
 public class Main{
 static long modularExponentiation(long x,long n,long M)
{
    if(n==0)
        return 1;
    else if(n%2 == 0)        //n is even
        return modularExponentiation((x*x)%M,n/2,M);
    else                             //n is odd
        return (x*modularExponentiation((x*x)%M,(n-1)/2,M))%M;

}
    public static void main(String[] args) throws Exception {
    Scanner in=new Scanner(System.in);
    int n=in.nextInt();
    long x=in.nextLong();
    ArrayList<Integer> array=new ArrayList<Integer>();
    while (n%2==0) 
        { 
            if(!array.contains(2))
            array.add(2); 
            n /= 2; 
        } 
  
     
        for (int i = 3; i <= Math.sqrt(n); i+= 2) 
        { 
            while (n%i == 0) 
            { 
                if(!array.contains(i))
                array.add(i); 
                n /= i; 
            } 
        } 
  
        if (n > 2) {
        if(!array.contains(n)) array.add(n);
        }
        long a[]=new long[(int)array.size()]; Arrays.fill(a,0);
            for(int i=0;i<array.size();i++)
            {
                long count=array.get(i);
                long num=x/count;
                while(num!=0) { 
                a[i]+=num; num=num/count;}
                
            }

            long ans=1;
            for(int i=0;i<a.length;i++)
            {
                long mul=1;
                long count=array.get(i);
                mul=modularExponentiation(count,a[i],1000000007);  
                ans=(ans*mul)%1000000007;
            }
            System.out.println(ans);
    
}
}