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.InputStreamReader;
import java.io.PrintWriter;
import java.lang.reflect.Array;
import java.util.*;

public class F {
    public static boolean is_prime(int a){
        for(int j = 2; j <= Math.sqrt(a); ++j){
            if(a % j == 0){
                return false;
            }
        }
        return true;
    }
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter writer = new PrintWriter(System.out);
        StringTokenizer st = new StringTokenizer(reader.readLine());
        long x = Long.parseLong(st.nextToken());
        long n = Long.parseLong(st.nextToken());
        ArrayList<Long> del = new ArrayList<>();
        for(int i = 2; i <= Math.sqrt(x); ++i){
            if(x % i == 0){
                if(is_prime(i)){
                    del.add((long)i);
                }
                if(x / i != i && is_prime((int)x / i)){
                    del.add(x / i);
                }
            }
        }
        if(is_prime((int)x)){
            del.add(x);
        }
        int MOD = 1000000007;
        long[][] arr = new long[del.size()][62];
        for(int j = 0; j < del.size(); ++j){
            long deli = del.get(j);
            arr[j][0] = deli;
            for(int i = 1; i < 62; ++i){
                arr[j][i] = arr[j][i - 1] * arr[j][i - 1];
                arr[j][i] %= MOD;
            }
        }
        long ans = 1;
        for(int j = 0; j < del.size(); ++j){
            long deli = del.get(j);
            long counter = 0;
            long cop = n;
            while(cop > 0){
                counter += cop / deli;
                cop /= deli;
            }
            char[] b = Long.toBinaryString(counter).toCharArray();
            for(int i = b.length - 1; i >= 0; --i){
                if(b[i] == '1'){
                    ans *= arr[j][b.length - 1 - i];
                    ans %= MOD;
                }
            }
        }
        System.out.println(ans);
    }
}