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.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;

/**
 * Created by pallavi on 8/1/16.
 */
public class D615 {
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        int m = Integer.parseInt(reader.readLine());
        String[] s = reader.readLine().split(" ");
        long[] ct = new long[200010];
        long n = 1;
        long mod = 1000000007;
        int[] p = new int[m];
        for (int i = 0; i < m; i++) {
            p[i] = Integer.parseInt(s[i]);
            ct[p[i]]++;
            n *= (long)p[i];
            n %= mod;
        }
        long d = 1;
        long sqn = 1;
        boolean ps = true;
        for (int i = 2; i < 200010; i++) {
            if (ct[i]>0) {
                if (ps && (ct[i]%2==1))  {
                    d *= (ct[i]+1)/2;
                    ps = false;
                } else {
                    d *= (ct[i]+1);
                }
                d %= (mod-1);
                for (int j = 0; j < ct[i]/2; j++) {
                    sqn *= (long)i;
                    sqn %= mod;
                }
            }
        }
        BigInteger exp = new BigInteger(String.valueOf(d));
        BigInteger MOD = new BigInteger(String.valueOf("1000000007"));
        BigInteger N;
        if (ps) {
            N = new BigInteger(String.valueOf(sqn));
        } else {
            N = new BigInteger(String.valueOf(n));
        }
        System.out.println(N.modPow(exp, MOD));
    }
}