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.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayDeque;
import java.util.Arrays;
import java.util.Scanner;

public class C_big {
    static boolean isPrime[];
    static int primes[];

    public static void main(String[] args) throws Exception {
        Scanner in = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        sieve((int) 1e5);
        int x = in.nextInt();
        long n = in.nextLong();
        ArrayDeque<Integer> pf = primeFactors(x);
        long ans = 1, mod = (long) 1e9 + 7;
        BigInteger N = BigInteger.valueOf(n);
        for (int p : pf) {
            BigInteger P = BigInteger.valueOf(p);
            for (BigInteger i = P; i.compareTo(N) <= 0; i = i.multiply(P)) {
                long d = n / i.longValue();
                ans = (ans * bigMod(p, d, mod)) % mod;
            }
        }

        pw.println(ans);

        pw.close();
    }

    static long bigMod(long a, long p, long m) { // returns (a^p) % m
        long res = 1 % m, x = a % m;
        while (p > 0) {
            if ((p & 1) > 0) res = (res * x) % m;
            x = (x * x) % m;
            p >>= 1;
        }
        return res;
    }

    static ArrayDeque<Integer> primeFactors(int n) {
        ArrayDeque<Integer> pf = new ArrayDeque<>();
        for (int p : primes) {
            if (n <= 1) break;
            if (n % p == 0) {
                pf.addLast(p);
                while (n % p == 0) n /= p;
            }
        }
        if (n > 1) pf.addLast(n);
        return pf;
    }

    static int[] sieve(int n) {
        isPrime = new boolean[n + 1];
        Arrays.fill(isPrime, true);
        isPrime[0] = isPrime[1] = false;
        for (int i = 4; i <= n; i += 2) isPrime[i] = false;
        for (int i = 3; i * i <= n; i += 2) {
            if (isPrime[i]) {
                for (int j = i * i; j <= n; j += 2 * i)
                    isPrime[j] = false;
            }
        }
        primes = new int[n + 1];
        int idx = 0;
        primes[idx++] = 2;
        for (int i = 3; i <= n; i += 2) {
            if (isPrime[i]) {
                primes[idx++] = i;
            }
        }
        return primes = Arrays.copyOf(primes, idx);
    }

    static void debug(Object... obj) {
        System.err.println(Arrays.deepToString(obj));
    }
}