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.BufferedReader;
import java.io.InputStreamReader;

public class Multipliers {
    static long mod = 1000000000 + 7;
    static int max = 200000;
    static long counts[] = new long[max + 5];
    static long multRight[] = new long[max + 5];
    static long multLeft[] = new long[max + 5];

    public static void main(String[] args) throws Exception {
        BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
        int m = Integer.parseInt(bf.readLine());
        String[] l = bf.readLine().split(" ");
        int curr;
        for (int i = 0; i < l.length; i++) {
            curr = Integer.parseInt(l[i]);
            counts[curr]++;
        }
        multLeft[0] = 1;
        for (int i = 1; i <= max; i++) {
            multLeft[i] = (counts[i] + 1);
            multLeft[i] *= multLeft[i - 1];
            multLeft[i] %= (mod - 1);
        }
        multRight[max + 1] = 1;
        for (int i = max; i >= 0; i--) {
            multRight[i] = (counts[i] + 1);
            multRight[i] *= multRight[i + 1];
            multRight[i] %= (mod - 1);
        }
        long powers, t, ans = 1;
        for (int i = 1; i <= max; i++) {
            if (counts[i] == 0)
                continue;
            powers = multLeft[i - 1] * multRight[i + 1];
            powers %= (mod - 1);
            powers *= (sum(counts[i]) % (mod - 1));
            powers %= (mod - 1);
            powers = binaryPow(i, powers);
            ans *= powers;
            ans %= mod;
        }
        System.out.println(ans);
    }

    private static long sum(long i) {
        return i * (i + 1) / 2;
    }

    static long binaryPow(long a, long q) {
        a %= mod;
        if (q == 0)
            return 1;
        return ((q % 2 == 1 ? a : 1) * binaryPow(a * a, q / 2)) % mod;
    }
}
