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 Hello {
    	public static TreeSet<Long> primeFactors(long n)
        {
    		TreeSet<Long> res = new TreeSet<Long>();
            // Print the number of 2s that divide n 
            while (n % 2 == 0) { 
                n /= 2;
                res.add((long) 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) { 
                    res.add((long)i);
                    n /= i; 
                }
            } 
      
            // This condition is to handle the case whien 
            // n is a prime number greater than 2 
            if (n > 2) 
                res.add(n);
            return res;
        }
    	  static long power(long x, long y, long p) 
    	    { 
    	        // Initialize result 
    	        long res = 1;      
    	         
    	        // Update x if it is more   
    	        // than or equal to p 
    	        x = x % p;  
    	      
    	        while (y > 0) 
    	        { 
    	            // If y is odd, multiply x 
    	            // with result 
    	            if((y & 1)==1) 
    	                res = (res * x) % p; 
    	      
    	            // y must be even now 
    	            // y = y / 2 
    	            y = y >> 1;  
    	            x = (x * x) % p;  
    	        } 
    	        return res; 
    	    } 
    	static long A = 1000000000+7;
    	public static void main(String[] args) {
    		Scanner sc = new Scanner(System.in);
    		long x = sc.nextLong();
    		long n = sc.nextLong();
    		TreeSet<Long> primes = primeFactors(x);
    		//System.out.println(primes);
    		long res = 1;
    		while(!primes.isEmpty()) {
    			long a = primes.pollFirst();
    			long t = a;
    			int c = 1;
    			while(a<=n) {
    				//System.out.println(">>"+res + " " +a);
    				res = (res*power(t,n/a,A))%A;
    				try {
    					 a = Math.multiplyExact(a,t);
    					} catch(ArithmeticException e) {
    					  break;
    					}
    				c++;
    			}
    		}
    		System.out.println(res);
    	}
    }