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.util.Scanner;

/*
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 * and open the template in the editor.
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/**
 *
 * @author hitarthk
 */
public class C_338_D {
    public static void main(String[] args) {
        Scanner in=new Scanner(System.in);
        int m=in.nextInt();
        long[] f=new long[200002];
        for(int i=1;i<=m;i++){
            f[in.nextInt()]++;
        }long mod=(long)1e9+7;
        long ans=1;
        long count=1;
        for(int i=2;i<=200000;i++){
            if(f[i]>0){
                ans=(modPow(ans, f[i]+1, mod)*modPow(i, (((f[i]*(f[i]+1))/2)%(mod-1)*count%(mod-1))%(mod-1), mod))%mod;
                count=(count*(f[i]+1))%(mod-1);
//                System.out.println(ans+" "+count);
            }
        }
        System.out.println(ans);
    }
    
    static long modPow(long b,long e,long mod){
        if(e==0){
            return 1;
        }
        else{
            long ans=modPow(b, e/2, mod);
            ans=(ans*ans)%mod;
            if(e%2!=0){
                ans=(ans*b)%mod;
            }
            return ans;
                    
        }
            
    }
    
}
