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 {

    long mod = (int) 1e9 + 7;

    long binpow(long a, long n) {
        long res = 1;
        while (n > 0) {
            if ((n & 1) > 0) {
                res *= a;
                res %= mod;
            }
            a *= a;
            a %= mod;
            n >>= 1;
        }
        return res;
    }

    public void solve() throws IOException {
        long x = nextInt(), n = nextLong();
        Map<Integer, Long> fact = new HashMap<>();
        Set<Integer> div = new HashSet<>();
        int t = (int) x;
        for (int d = 2; d * d <= t; d++) {
            while (t % d == 0) {
                fact.putIfAbsent(d, 0L);
                div.add(d);
                t /= d;
            }
        }
        if (t > 1) {
            fact.putIfAbsent(t, 0L);
            div.add(t);
        }
        for (Integer v : div) {
            BigInteger cur = BigInteger.valueOf(v);
            while (cur.compareTo(BigInteger.valueOf(n)) <= 0) {
                fact.put(v, fact.get(v) + n / cur.longValue());
                cur = cur.multiply(BigInteger.valueOf(v));
            }
        }
        BigInteger ans = BigInteger.ONE;
        for (Map.Entry<Integer, Long> e : fact.entrySet()) {
            if (e.getValue() > 0) {
                ans = ans.multiply(BigInteger.valueOf(binpow(e.getKey(), e.getValue())));
                ans = ans.mod(BigInteger.valueOf(mod));
            }
        }
        out.print(ans);
    }

    public void run() {
        try {
            br = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);

            solve();

            out.close();
        } catch (IOException e) {
            e.printStackTrace();
            System.exit(1);
        }
    }

    BufferedReader br;
    StringTokenizer in;
    PrintWriter out;

    public String nextToken() throws IOException {
        while (in == null || !in.hasMoreTokens()) {
            in = new StringTokenizer(br.readLine());
        }
        return in.nextToken();
    }

    public int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    public double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    public long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    public int[] nextArr(int n) throws IOException {
        int[] res = new int[n];
        for (int i = 0; i < n; i++) {
            res[i] = nextInt();
        }
        return res;
    }

    public static void main(String[] args) throws IOException {
        Locale.setDefault(Locale.US);
        new C().run();
    }
}