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.*;
import java.util.*;


public class test {
    int INF = (int)1e9;
    long MOD = 1000000007;
    
    
    long pow(long a, long n) {
        long rs = 1;
        while(n>0) {
            if(n%2==1) {
                rs *= a;
                rs %= MOD;
            }
            a *= a;
            a %= MOD;
            n /= 2;
        }
        return rs;
    }
    
    void solve(InputReader in, PrintWriter out) throws IOException {
        int n = in.nextInt();
        int[] c = new int[200010];
        for(int i=0; i<n; i++) {
            int x = in.nextInt();
            c[x]++;
        }
        long rs = 1;
        long d = 1;
        for(int i=1; i<c.length; i++) if(c[i]>0) {
            long x = c[i];
            long v = pow(i, x*(x+1)/2);
            rs = pow(rs, x+1);
            rs *= pow(v, d);
            rs %= MOD;
            d = d*(x+1) % (MOD-1);
        }
        out.println(rs);
    }
    
    
    public static void main(String[] args) throws IOException {
        if(args.length>0 && args[0].equalsIgnoreCase("d")) {
            DEBUG_FLAG = true;
        }
        InputReader in = new InputReader();
        PrintWriter out = new PrintWriter(System.out);
        int t = 1;//in.nextInt();
        long start = System.nanoTime();
        while(t-->0) {
            new test().solve(in, out);
        }
        long end = System.nanoTime();
        debug("\nTime: " + (end-start)/1e6 + " \n\n");
        out.close();
    }
    
    static boolean DEBUG_FLAG = false;
    static void debug(String s) {
        if(DEBUG_FLAG) System.out.print(s);
    }
    
    public static void show(Object... o) {
        System.out.println(Arrays.deepToString(o));
    }
    
    static class InputReader {
        static BufferedReader br;
        static StringTokenizer st;
    
        public InputReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        
        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        
        int nextInt() {
            return Integer.parseInt(next());
        }
        
        long nextLong() {
            return Long.parseLong(next());
        }
        
        double nextDouble() {
            return Double.parseDouble(next());
        }
    }
}