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 MAX = 2e6 + 10;
const int MAXN = 1e4 + 10;
const int MOD = 1e9 + 7;
const int inf = 1e9;
const double pi = acos(-1.0);
const double eps = 1e-6;
long long m, p;
long long res, tmp, temp;
long long cnt[MAX];
long long Pow(long long b, long long p) {
  if (!p) return 1;
  long long ret = Pow(b, p >> 1);
  ret = (ret * ret) % MOD;
  if (p & 1) ret = (ret * b) % MOD;
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  res = 1;
  cin >> m;
  for (int i = 0; i < m; i++) {
    cin >> p;
    cnt[p]++;
  }
  p = 1;
  for (int i = 2; i <= 2e6; i++) {
    tmp = 1;
    for (int j = 1; j <= cnt[i]; j++) {
      temp = (res * Pow(i, (p * j) % (MOD - 1))) % MOD;
      tmp = (tmp * temp) % MOD;
    }
    res = (res * tmp) % MOD;
    p = (p * (cnt[i] + 1)) % (MOD - 1);
  }
  cout << res;
  return 0;
}
