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 = 1e9 + 7;
long long power(long long a, int e, int mod) {
  long long result = 1;
  while (e) {
    if (e & 1) {
      result = (result * a) % mod;
    }
    a = (a * a) % mod;
    e >>= 1;
  }
  return result;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.precision(10);
  cout << fixed;
  int n;
  cin >> n;
  vector<int> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  sort(v.begin(), v.end());
  long long prod = 1;
  long long count = 0;
  long long nr = 1;
  for (int i = 0; i < n; i++) {
    count++;
    if (i == n - 1 || v[i] != v[i + 1]) {
      prod = power(prod, count + 1, MOD);
      long long p = ((1 + count) * count / 2) % (MOD - 1);
      p = (p * nr) % (MOD - 1);
      long long t = power(v[i], p, MOD);
      prod = (prod * t) % MOD;
      nr = (nr * (count + 1)) % (MOD - 1);
      count = 0;
    }
  }
  cout << prod << endl;
  return 0;
}
