Ayrat has number n, represented as it's prime factorization pi of size m, i.e. n = p1·p2·...·pm. Ayrat got secret information that that the product of all divisors of n taken modulo 109 + 7 is the password to the secret data base. Now he wants to calculate this value.

Input

The first line of the input contains a single integer m (1 ≤ m ≤ 200 000) — the number of primes in factorization of n. 

The second line contains m primes numbers pi (2 ≤ pi ≤ 200 000).

Output

Print one integer — the product of all divisors of n modulo 109 + 7.

Examples

Input

2
2 3


Output

36


Input

3
2 3 2


Output

1728

Note

In the first sample n = 2·3 = 6. The divisors of 6 are 1, 2, 3 and 6, their product is equal to 1·2·3·6 = 36.

In the second sample 2·3·2 = 12. The divisors of 12 are 1, 2, 3, 4, 6 and 12. 1·2·3·4·6·12 = 1728.
import java.io.IOException;
import java.util.*;

public class Main {

    public static final long MOD = 1000000007;

    public static void main(String[] args) throws IOException {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        long[] p = new long[n];
        for (int i = 0; i < n; ++i) {
            p[i] = in.nextInt();
        }
        Arrays.sort(p);
        List<Integer> m = new ArrayList<>();
        int c = 1;
        for (int i = 1; i < n; ++i) {
            if (p[i] != p[i - 1]) {
                m.add(c);
                c = 0;
            }
            ++c;
        }
        m.add(c);
        System.out.println(solve(p, m));

    }

    private static long solve(long[] p, List<Integer> m) {
        long res = 1;

        for (long aP : p) {
            res = (res * aP) % MOD;
        }

        boolean ok = false;
        for (Integer mI : m) {
            if ((mI & 1) != 0) {
                ok = true;
                break;
            }
        }
        if (!ok) {
            res = 1;
            long prevA = 0;
            int cnt = 0;
            int index = -1;
            for (long aP : p) {
                if (aP != prevA) {
                    cnt = 0;
                    prevA = aP;
                    ++index;
                }
                if (cnt < m.get(index) / 2) {
                    res = (res * aP) % MOD;
                    ++cnt;
                }
            }
        }

        long prod = 1;
        boolean halved = !ok;
        for (int i = 0; i < m.size(); ++i) {
            long pw = (m.get(i) + 1);
            if ((pw & 1) == 0 && !halved) {
                pw /= 2;
                halved = true;
            }
            res = binPow(res, pw);
        }
        return res;
    }

    private static long binPow(long a, long i) {
        if (i == 0) return 1;
        long res = 1;
        while (i > 0) {
            if ((i & 1) != 0) {
                res = (res * a) % MOD;
            }
            a = (a * a) % MOD;
            i >>= 1;
        }
        return res;
    }
}