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;
long long int binary_exp(long long int a, long long int n) {
  long long int i = 1;
  while (n > 0) {
    if (n % 2 == 1) i = (i * a) % 1000000007;
    a = (a * a) % 1000000007;
    n /= 2;
  }
  return i;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  srand(time(NULL));
  ;
  map<long long int, long long int> m;
  long long int n, i, j, k = 0, ans = 1, b = 1, c = 1;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> j;
    m[j]++;
  }
  map<long long int, long long int>::iterator it;
  for (it = m.begin(); it != m.end(); it++) {
    if (it->second % 2 == 1) k++;
  }
  if (k == 0) {
    for (it = m.begin(); it != m.end(); it++)
      ans = (ans * binary_exp(it->first, it->second / 2)) % 1000000007;
    c = 0;
  } else {
    for (it = m.begin(); it != m.end(); it++)
      ans = (ans * binary_exp(it->first, it->second)) % 1000000007;
  }
  for (it = m.begin(); it != m.end(); it++) {
    if (c && it->second % 2 == 1) {
      c = 0;
      ans = binary_exp(ans, (it->second + 1) / 2);
    } else
      ans = binary_exp(ans, it->second + 1);
  }
  cout << ans;
}
