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.*;

public class c589
{
	public static void main(String[] args) throws IOException
	{new c589();}
	
	FastScanner in = new FastScanner(System.in);
	PrintWriter out = new PrintWriter(System.out);
	
	int x, max = 100_000;
	long n, ans, mod = (long)1e9+7;
	int[] pr, div_x;
	boolean[] isp;
	
	c589() throws IOException
	{
		x = in.nextInt();
		n = in.nextLong();
		
		pr = new int[max];
		isp = new boolean[max+1];
		for (int i = 2; i*i <= max; i++)
			if (!isp[i])
				for (int j = 2*i; j <= max; j += i)
					isp[j] = true;
		for (int i = 2; i <= max; i++)
			if (!isp[i])
				pr[++pr[0]] = i;
		
		div_x = new int[20];
		for (int i = 1; i <= pr[0]; i++)
		{
			if (x%pr[i] == 0)
				div_x[++div_x[0]] = pr[i];
			while (x%pr[i] == 0)
				x /= pr[i];
		}
		if (x != 1)
			div_x[++div_x[0]] = x;
		
		ans = 1L;
		for (int i = 1; i <= div_x[0]; i++)
		{
			long e = 0, nn = n;
			while (nn > 0)
			{
				nn /= div_x[i];
				e += nn;
			}
			ans *= exp(div_x[i], e);
			ans %= mod;
		}
		
		out.println(ans);
		out.close();	
	}
	
	long exp(long a, long b)
	{
		if (b == 0)
			return 1L;
		if (b == 1)
			return a;
		long ans = exp(a, b/2);
		ans *= ans;
		ans %= mod;
		if ((b & 1) == 0)
			return ans;
		return (a*ans) % mod;
	}
	
	class FastScanner {
		BufferedReader br;
		StringTokenizer st;
		public FastScanner(InputStream i) {
		    br = new BufferedReader(new InputStreamReader(i));
		    st = new StringTokenizer("");
		}		
		public String next() throws IOException {
		    if(st.hasMoreTokens())
		        return st.nextToken();
		    else
		        st = new StringTokenizer(br.readLine());
		    return next();
		}
		public int nextInt() throws IOException {
		    return Integer.parseInt(next());
		}
		public int[] intarr(int n) throws IOException {
			int[] res = new int[n];
			for (int i = 0; i < n; i++)
				res[i] = nextInt();
			return res;
		}
	    public long nextLong() throws IOException {
		    return Long.parseLong(next());
		}
		public double nextDouble() throws IOException {
		    return Double.parseDouble(next());
		}
    }
}