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.ArrayList;
import java.util.Scanner;

public class Main {

    private static long module = 1000000007;

    public static void main(String[] args) {
	    Scanner in = new Scanner(System.in);
	    long x = in.nextLong();
	    long n = in.nextLong();
	    long answer = 1;



	    ArrayList<Long> primeX = getPrimeFactors(x);
        long[] power = new long[primeX.size()];
        for (int i = 0; i < primeX.size() ; i++) {
            power[i] = 0;
        }

        for (int i = 0; i < power.length ; i++) {
            power[i] = getPower(n, primeX.get(i));
        }

        for (int i = 0; i < power.length ; i++) {
            String strPower = Long.toBinaryString(power[i]);
            String[] splitPower = strPower.split("");
            long[] masBin = new long[splitPower.length];
            int k = 0;
            for (int j = splitPower.length - 1; j >=0 ; j--) {
                masBin[k] = Long.parseLong(splitPower[j]);
                k++;
            }

            answer *= binPow(primeX.get(i), masBin);
            answer = answer % module;

        }

        System.out.println(answer);
    }

    private static long binPow(long number, long[] masBinaryNumber){
        long r =1;
        long n = number;
        long[] masNumberBinaryM = masBinaryNumber;
        for (int i = 0; i < masNumberBinaryM.length ; i++) {
            if(masNumberBinaryM[i]==1){
                r = (r*n)%module;
                n = (n*n)%module;
            }
            else{
                r = r%module;
                n = (n*n)%module;
            }
        }
        return r;
    }

    public static long getPower(long x, long primeNumber){
        long number = x;
        long prime = primeNumber;
        long sum = 0;

        while(number > 0){
            long p = number /prime;
            sum += p;
            number = p;
        }

        return sum;
    }


    public static ArrayList<Long> getPrimeFactors(long num){
        ArrayList<Long> prime = new ArrayList<>();
        int flag = 0;
        long b = num;
        long number = num;
        for (long i = 2; i * i <= b; i++) {
            while (number % i == 0) {
                number /= i;
                if (number == 1) {
                    for (int j = 0; j < prime.size() ; j++) {
                        if(prime.get(j) == i){
                            flag++;
                        }
                    }
                    if(flag == 0){
                        prime.add((i));
                        break;
                    }
                }
                for (int j = 0; j < prime.size() ; j++) {
                    if(prime.get(j) == i){
                        flag++;
                    }
                }
                if(flag == 0){
                    prime.add((i));
                }
                flag = 0;
            }
        }
        if (number != 1) {
            for (int j = 0; j < prime.size() ; j++) {
                if(prime.get(j) == number){
                    flag++;
                }
            }
            if(flag == 0){
                prime.add((number));
            }
        }
        return prime;
    }
}
