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.*;
import java.io.*;
import java.math.*;
import java.lang.*;
public class Main {
	public static int maxn = 1005;
	public static int fac[] = new int[maxn],cnt=0;
	public static void getfac(int x){
		for(int i=2;i*i<=x;i++){
			if(x%i==0){
				fac[++cnt]=i;
				while(x%i==0){
					x/=i;
				}
			}
		}
		if(x>1) fac[++cnt]=x;
	}
	public static void main(String[] args) {
		InputReader in = new InputReader();
		PrintWriter out = new PrintWriter(System.out);
		
		BigInteger Mod = BigInteger.valueOf(1000000007);
		int x=in.nextInt(); 
		BigInteger n=in.nextBigInteger();
		getfac(x);
		
		BigInteger ans=BigInteger.ONE;
		BigInteger num1=BigInteger.ZERO,num2=BigInteger.ZERO;
		
		for(int i=1;i<=cnt;i++){
			BigInteger p = BigInteger.valueOf(fac[i]);
			BigInteger pre = p, now = p.multiply(p);
			for(int k=1;k<=60;k++){
				num1 = n.divide(pre);
				num2 = n.divide(now);
				ans = ans.multiply(pre.modPow(num1.subtract(num2), Mod));
				ans = ans.mod(Mod);
				pre=now;
				now=now.multiply(p);
			}
		}
		out.println(ans);
		out.flush();
	}
}


class InputReader {
	BufferedReader buf;
	StringTokenizer tok;
	InputReader() {
		buf = new BufferedReader(new InputStreamReader(System.in));
	}
	boolean hasNext() {
		while (tok == null || !tok.hasMoreElements()) {
			try {
				tok = new StringTokenizer(buf.readLine());
			} catch (Exception e) {
				return false;
			}
		}
		return true;
	}
	String next() {
		if (hasNext())
			return tok.nextToken();
		return null;
	}
	int nextInt() {
		return Integer.parseInt(next());
	}
	long nextLong() {
		return Long.parseLong(next());
	}
	double nextDouble() {
		return Double.parseDouble(next());
	}
	BigInteger nextBigInteger() {
		return new BigInteger(next());
	}
	BigDecimal nextBigDecimal() {
		return new BigDecimal(next());
	}
}