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.math.*;
import java.security.KeyStore.Entry;
import java.util.*;

public class SPOJANDCC {
	private InputStream is;
	private PrintWriter out;

	long mod = (long) 1e9 + 7;

	long pow(long x, long y, long m) {
		if (y == 0)
			return 1;
		long k = pow(x, y / 2, m);
		if (y % 2 == 0)
			return (k * k) % m;
		else
			return (((k * k) % m) * x) % m;
	}

	long mul(long a, long b) {
		return (a * b) % mod;
	}

	void solve() {
		int n = ni();
		long arr[] = new long[n];
		for (int i = 0; i < n; i++)
			arr[i] = ni();
		Arrays.sort(arr);
		long ans = 1, b = arr[0];
		long cnt = 0, t = 1;
		for (int i = 1; i < n; i++) {
			if (arr[i] == arr[i - 1])
				b = mul(b, pow(arr[i], ++t, mod));
			else {
				ans = pow(ans, t + 1, mod);
				b = pow(b, cnt + 1, mod);
				ans = mul(ans, b);
				b = arr[i];
				cnt = (cnt + 1) * (t + 1) - 1;
				cnt %= (mod - 1);
				t = 1;
			}
		}
		ans = pow(ans, t + 1, mod);
		b = pow(b, cnt + 1, mod);
		ans = mul(ans, b);
		out.println(ans);
	}

	void soln() {
		is = System.in;
		out = new PrintWriter(System.out);
		solve();
		out.flush();
	}

	public static void main(String[] args) throws Exception {
		new SPOJANDCC().soln();
	}

	// To Get Input
	// Some Buffer Methods
	private byte[] inbuf = new byte[1024];
	public int lenbuf = 0, ptrbuf = 0;

	private int readByte() {
		if (lenbuf == -1)
			throw new InputMismatchException();
		if (ptrbuf >= lenbuf) {
			ptrbuf = 0;
			try {
				lenbuf = is.read(inbuf);
			} catch (IOException e) {
				throw new InputMismatchException();
			}
			if (lenbuf <= 0)
				return -1;
		}
		return inbuf[ptrbuf++];
	}

	private boolean isSpaceChar(int c) {
		return !(c >= 33 && c <= 126);
	}

	private int skip() {
		int b;
		while ((b = readByte()) != -1 && isSpaceChar(b))
			;
		return b;
	}

	private double nd() {
		return Double.parseDouble(ns());
	}

	private char nc() {
		return (char) skip();
	}

	private String ns() {
		int b = skip();
		StringBuilder sb = new StringBuilder();
		while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != '
									// ')
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}

	private char[] ns(int n) {
		char[] buf = new char[n];
		int b = skip(), p = 0;
		while (p < n && !(isSpaceChar(b))) {
			buf[p++] = (char) b;
			b = readByte();
		}
		return n == p ? buf : Arrays.copyOf(buf, p);
	}

	private char[][] nm(int n, int m) {
		char[][] map = new char[n][];
		for (int i = 0; i < n; i++)
			map[i] = ns(m);
		return map;
	}

	private int[] na(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = ni();
		return a;
	}

	private int ni() {
		int num = 0, b;
		boolean minus = false;
		while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
			;
		if (b == '-') {
			minus = true;
			b = readByte();
		}

		while (true) {
			if (b >= '0' && b <= '9') {
				num = num * 10 + (b - '0');
			} else {
				return minus ? -num : num;
			}
			b = readByte();
		}
	}

	private long nl() {
		long num = 0;
		int b;
		boolean minus = false;
		while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
			;
		if (b == '-') {
			minus = true;
			b = readByte();
		}

		while (true) {
			if (b >= '0' && b <= '9') {
				num = num * 10 + (b - '0');
			} else {
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
}