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;
inline long long int power(long long int a, long long int b, long long int c) {
  long long int ans = 1;
  while (b > 0) {
    if (b & 1) {
      ans = (ans * a) % c;
    }
    a = (a * a) % c;
    b = b >> 1;
  }
  return ans;
}
long long int v[200001];
int main() {
  memset(v, 0, sizeof(0));
  long long int m, t;
  cin >> m;
  for (long long int i = 0; i < m; ++i) {
    cin >> t;
    v[t]++;
  }
  long long int n = 1;
  long long int ans = 1;
  for (long long int i = 1; i <= 200000; ++i) {
    if (v[i] != 0) {
      t = power(i, (v[i] * (v[i] + 1)) / 2, 1000000007);
      ans = (power(ans, v[i] + 1, 1000000007) * power(t, n, 1000000007)) %
            1000000007;
      n = (n * (v[i] + 1)) % (1000000007 - 1);
    }
  }
  cout << ans << endl;
  return 0;
}
