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 final class Main
{ 
   public static HashSet<Long> g(long n) 
    { 
        // Print the number of 2s that divide n
        HashSet<Long> set=new HashSet();
        while (n%2==0) 
        { 
            set.add(Long.valueOf(2)); 
            n /= 2; 
        } 
  
        // n must be odd at this point.  So we can 
        // skip one element (Note i = i +2) 
        for (long i = 3; i <= Math.sqrt(n); i+= 2) 
        { 
            // While i divides n, print i and divide n 
            while (n%i == 0) 
            { 
                set.add((long)i); 
                n /= i; 
            } 
        } 
  
        // This condition is to handle the case whien 
        // n is a prime number greater than 2 
        if (n > 2) 
            set.add((long)n);
            return set;
    }   
    // Driver method 
    public static void main(String args[]) 
    { Scanner sc=new Scanner(System.in);
        
         
        long x=sc.nextLong();
        long n=sc.nextLong();
        HashSet <Long> P = g(x);
        Iterator itr=P.iterator();
        long mod=1000000007;
        long ans=Long.valueOf(1);
       while(itr.hasNext())
       {
          x=(Long)itr.next();
          long k=0;
          long z=n;
          while(z>0)
          {
              k+=(z/x);
              z/=x;
          }
          ans*=powm(x,k,mod);
          ans%=mod;
       }
      System.out.println(ans);
    }
    public static long pow(long x,long y)
	{
	    if(y==0)
	    return 1;
	    long z=pow(x,y/Long.valueOf(2));
	    z=(z*z);
	    if(y%2==1)
	    {
	        z=(x*z);
	    }
	    return z;
	}
	public static long powm(long x,long y,long mod)
	{
	    x%=mod;
	    if(y==0)
	    return 1;
	    long z=powm(x,y/Long.valueOf(2),mod);
	    z=((z%mod)*(z%mod))%mod;;
	    if(y%2==1)
	    {
	        z=((x%mod)*z)%mod;
	    }
	    return z;
	}
} 

