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.math.BigInteger;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Scanner;
import java.util.Set;

public class PrimesAndMultiplication {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);
    int x = scanner.nextInt();
    long n = scanner.nextLong();
    List<Integer> primesLessThanTenToPower5 = getPrimesLessThanTenToPower5();
    Set<Integer> primeFactorsOfX = primeFactors(x, primesLessThanTenToPower5);

    long result = 1;
    for (Integer prime: primeFactorsOfX) {
      result = (result * getPowerOfPrimeInFactorialOf(prime, n)) % RESULT_MODULO;
    }

    System.out.println(result);
  }

  private static long getPowerOfPrimeInFactorialOf(int prime, long n) {
    long power = 0;
    long curDividor = prime;

    while (n >= curDividor) {
      power += (n / curDividor);
      if ((n / prime) < curDividor) {
        break;
      }
      curDividor *= prime;
    }

    return BigInteger
        .valueOf(prime)
        .modPow(BigInteger.valueOf(power), BigInteger.valueOf(RESULT_MODULO))
        .longValue();
  }

  // Returns a map from prime to its power in the factorization
  private static Set<Integer> primeFactors(
      int x, List<Integer> primesLessThanTenToPower5) {
    Set<Integer> factors = new HashSet<>();
    for (Integer prime: primesLessThanTenToPower5) {
      int count = 0;
      while (x % prime == 0) {
        x = x / prime;
        count++;
      }
      if (count != 0) {
        factors.add(prime);
      }
    }
    if (x > 1) {
      factors.add(x);
    }
    return factors;
  }

  private static List<Integer> getPrimesLessThanTenToPower5() {
    List<Integer> result = new ArrayList<>();
    // Value at index i is 1 if and only if (i + 1) is prime. Otherwise, -1.
    // 0 implies unknown.
    int[] sieveForTenToPower5 = new int[TEN_TO_POWER_5];
    final int unknown = 0;
    final int prime = 1;
    final int non_prime = -1;
    sieveForTenToPower5[0] = non_prime;
    int i = 1;
    while (i < TEN_TO_POWER_5) {
      if (sieveForTenToPower5[i] != unknown) {
        i++;
        continue;
      }
      sieveForTenToPower5[i] = prime;
      result.add(i + 1);
      int base = (i + 1);
      int multiple = base << 1;
      while (multiple <= TEN_TO_POWER_5) {
        sieveForTenToPower5[multiple - 1] = non_prime;
        multiple += base;
      }
      i++;
    }
    return result;
  }

  private static final int TEN_TO_POWER_5 = 100000;
  private static final int RESULT_MODULO = 1000000007;
}
