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;
const int mod = 1000000007;
int power(int n, int exp) {
  if (exp == 0) return 1;
  int temp = power(n, exp / 2);
  if (exp % 2 == 0)
    return (temp * 1LL * temp) % mod;
  else
    return (((temp * 1LL * temp) % mod) * 1LL * n) % mod;
}
int main(int argc, char *argv[]) {
  ios::sync_with_stdio(false);
  int m;
  cin >> m;
  int p, n = 1;
  map<int, int> times;
  for (int i = 0; i < m; i++) {
    cin >> p;
    times[p]++;
    n = (n * 1LL * p) % mod;
  }
  int divisors = 1;
  bool tworemaining = true;
  for (map<int, int>::iterator iter = times.begin(); iter != times.end();
       iter++) {
    int multiplier = (1 + iter->second);
    if (multiplier % 2 == 0 && tworemaining) {
      multiplier /= 2;
      tworemaining = false;
    }
    divisors = (divisors * 1LL * multiplier) % (mod - 1);
  }
  if (tworemaining) {
    int tempn = 1;
    for (map<int, int>::iterator iter = times.begin(); iter != times.end();
         iter++) {
      for (int i = 0; i < iter->second / 2; i++) {
        tempn = (tempn * 1LL * iter->first) % mod;
      }
    }
    n = tempn;
  }
  cout << power(n, divisors) << '\n';
}
