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.
#include <bits/stdc++.h>
using namespace std;
int l[200010], r[200010];
const int MOD = 1000000007;
const int MOD2 = 1000000006;
int p[200010];
int exp(int a, int b, int c) {
  int ans = 1, num = a;
  while (b > 0) {
    if (b & 1) {
      ans = (long long)ans * num % c;
    }
    num = (long long)num * num % c;
    b >>= 1;
  }
  return ans;
}
int main() {
  int m;
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    int x;
    scanf("%d", &x);
    p[x]++;
  }
  l[0] = 1;
  for (int i = 1; i <= 200001; i++) {
    l[i] = (long long)l[i - 1] * (p[i] + 1) % MOD2;
  }
  r[200001] = 1;
  for (int i = 200000; i >= 0; i--) {
    r[i] = (long long)r[i + 1] * (p[i] + 1) % MOD2;
  }
  int res = 1;
  for (int i = 2; i <= 200000; i++) {
    long long mult = (long long)l[i - 1] * r[i + 1] % MOD2;
    for (int j = 0; j < p[i]; j++) {
      long long tmp = (long long)mult * (p[i] - j) % MOD2;
      res = (long long)res * exp(i, tmp, MOD) % MOD;
    }
  }
  printf("%d\n", res);
  return 0;
}
