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.*;
import java.math.BigInteger;
import java.util.*;

public class C {
  private static final BigInteger TWO = new BigInteger("2");
  private static final BigInteger MOD = new BigInteger("1000000007");
  
  public static void main(String[] args) throws Exception {
    BufferedReader in = new BufferedReader(new InputStreamReader(System.in));

    StringTokenizer st = new StringTokenizer(in.readLine());
    int x = Integer.parseInt(st.nextToken());
    BigInteger n = new BigInteger(st.nextToken());

    System.out.println(solve(x, n));

    in.close();
  }

  private static BigInteger solve(int x, BigInteger n) {
    List<Integer> primes = findPrimes(x);

    BigInteger result = BigInteger.ONE;

    for (int p : primes) {
      BigInteger prime = new BigInteger(Integer.toString(p));
      BigInteger numPrime = BigInteger.ZERO;

      for (BigInteger i = prime; i.compareTo(n) <= 0; i = i.multiply(prime)) {
        numPrime = numPrime.add(n.divide(i));
      }

      BigInteger power = modPower(prime, numPrime);

      result = result.multiply(power).mod(MOD);
    }

    return result.mod(MOD);
  }
  
  private static List<Integer> findPrimes(int x) {
    List<Integer> primes = new ArrayList<>();

    int i = 2;
    while (i * i <= x) {
      if (x % i == 0) {
        primes.add(i);

        while (x % i == 0) {
          x /= i;
        }
      }

      i++;
    }

    if (x > 1) {
      primes.add(x);
    }

    return primes;
  }

  private static BigInteger modPower(BigInteger prime, BigInteger numPrime) {
    if (numPrime == BigInteger.ZERO) {
      return BigInteger.ONE;
    }

    if (numPrime.mod(TWO) == BigInteger.ZERO) {
      BigInteger result = modPower(prime, numPrime.divide(TWO)).mod(MOD);

      return result.multiply(result).mod(MOD);
    }

    return modPower(prime, numPrime.subtract(BigInteger.ONE))
            .mod(MOD)
            .multiply(prime)
            .mod(MOD);
  }
}
