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 org.omg.CORBA.INTERNAL;

import java.util.*;

public class Main {
    private static Scanner scanner;
    private static final int MODNUM = 1000 * 1000 * 1000 + 7;
    private static long x, n;
    private static ArrayList<Long> roots;

    private static void init () {
        scanner = new Scanner(System.in);
        x = scanner.nextLong();
        n = scanner.nextLong();
        roots = new ArrayList<Long>();
    }

    private static ArrayList<Long> findRoots (long number) {
        ArrayList<Long> rootsOfNumber = new ArrayList<Long>();

        long candidate = 2;
        while ((number > 1) && (candidate * candidate <= number)) {
            if (number % candidate == 0) {
                rootsOfNumber.add(candidate);
                while (number % candidate == 0)
                    number /= candidate;
            }

            ++candidate;
        }

        if (number > 1)
            rootsOfNumber.add(number);

        return rootsOfNumber;
    }

    private static long powerOfRootInAnswer (long root) {
        long result = 0;
        /*
        100 is always more than the maximum power.
        GoodNums are numbers which have exactly that amount of root.
         */
        long[] numOfGoodNums = new long[100];

        long poweredRoot = 1;
        int maxPower = 0;
        while (poweredRoot <= n / root) {
            poweredRoot *= root;
            numOfGoodNums[++maxPower] = n / poweredRoot;
            numOfGoodNums[maxPower - 1] -= numOfGoodNums[maxPower];
        }

        for (int i = 1; i <= maxPower; ++i)
            result += numOfGoodNums[i] * i;

        return result;
    }

    private static long power (long base, long exponent) {
        if (exponent == 0)
            return 1;
        if (exponent == 1)
            return base % MODNUM;

        long halfOfExponent = power(base, exponent / 2);

        long result = (halfOfExponent * halfOfExponent) % MODNUM;
        if (exponent % 2 == 1)
            result = (result * base) % MODNUM;

        return result;
    }

    private static long findAnswer () {
        long answer = 1;

        roots = findRoots (x);

        for (long root : roots) {
            answer *= power(root, powerOfRootInAnswer(root));
            answer %= MODNUM;
        }

        return answer;
    }

    public static void main (String[] args) {
        init();
        System.out.println(findAnswer());
    }
}