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;
using ll = long long;
const int MOD = 1e9 + 7;
const int N = 2 * 1e5 + 1;
int power(int x, int y) {
  int res = 1;
  while (y) {
    if (y % 2) res = (res * 1LL * x) % MOD;
    x = (x * 1LL * x) % MOD;
    y /= 2;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int m;
  cin >> m;
  int d = 1, prod = 1;
  vector<int> freq(N, 0);
  for (int i = 0; i < m; ++i) {
    int p;
    cin >> p;
    ++freq[p];
  }
  for (int i = 2; i < N; ++i) {
    if (freq[i] == 0) continue;
    int x = i;
    int k = freq[i];
    int exp = ((k * 1LL * (k + 1)) / 2LL) % (MOD - 1);
    int fp = power(x, exp);
    prod = (power(prod, k + 1) * 1LL * power(fp, d)) % MOD;
    d = (d * 1LL * (k + 1)) % (MOD - 1);
  }
  cout << prod << "\n";
}
