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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.HashSet;
import java.util.StringTokenizer;

public class ProblemC {

	public static void main(String[] args) {
		InputStream in = System.in;
		InputReader scan = new InputReader(in);
		int x = scan.nextInt();
		long n = scan.nextLong();
		HashSet<Integer> prmList = primeFactors(x);
		int mod = 1000000007;
		int res = 1;
		for(int prime:prmList) {
			long exp = getExponent(n,prime);
			int mul = fastModularExponent(prime, exp, mod);
			res = (int)(((long)res*mul)%mod);
		}
		System.out.println(res);
	}
	static long getExponent(long n,int num) {
		long exp = 0;
		while(n>=num) {
			exp += (n/num);
			n/=num;
		}
		return exp;
	}
	static int fastModularExponent(int base,long exponent,int mod) {
		int result = 1;
		//update base if it is >= mod
		base = base%mod;
		while(exponent>0) {
			if((exponent&1)==1) {
				//Converting to long to avoid overflow during multiplication
				result=(int)(((long)result*base)%mod);
			}
			exponent>>=1;
			base=(int)(((long)base*base)%mod);
		}
		return result;
	}
	static HashSet<Integer> primeFactors(int n) {
		HashSet<Integer> prmList = new HashSet<>();
		// Print the number of 2s that divide n
		if(n%2==0) {
			prmList.add(2);
		}
		while (n % 2 == 0) {
			n /= 2;
		}

		// n must be odd at this point. So we can
		// skip one element (Note i = i +2)
		for (int i = 3; i <= Math.sqrt(n); i += 2) {
			// While i divides n, print i and divide n
			while (n % i == 0) {
				n /= i;
				prmList.add(i);
			}
		}

		// This condition is to handle the case whien
		// n is a prime number greater than 2
		if (n > 2) prmList.add(n);
		return prmList;
	}
	static class InputReader {
		BufferedReader br;
		StringTokenizer st;
		
		InputReader(InputStream in) {
			br = new BufferedReader(new InputStreamReader(in));
		}
		
		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException ioe) {
					ioe.printStackTrace();
				}
			}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		public long nextLong() {
			return Long.parseLong(next());
		}
	}
}