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 maxn = 210000, mod = 1e9 + 7;
int a[maxn];
long long int pw(long long int a, int p) {
  if (p == 0) return 1;
  if (p == 1) return a % mod;
  long long int rt = pw(a, p / 2);
  rt = (rt * rt) % mod;
  if (p % 2) rt = (rt * a) % mod;
  return rt;
}
map<int, int> mp;
int main() {
  int m = 0;
  bool flag = false;
  long long int ans = 1;
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) {
    scanf("%d", &a[i]);
    mp[a[i]]++;
    ans = (ans * a[i]) % mod;
  }
  bool dv2 = false;
  for (auto p : mp) {
    if (p.second % 2) flag = true;
  }
  if (flag) {
    for (auto p : mp) {
      if (!dv2 && p.second % 2) {
        ans = pw(ans, (p.second + 1) / 2);
        dv2 = true;
      } else {
        ans = pw(ans, (p.second + 1));
      }
    }
  } else {
    for (auto p : mp) {
      if (!dv2) {
        ans = pw(ans, (p.second + 1) / 2);
        dv2 = true;
        for (auto p : mp) {
          ans = (ans * pw(p.first, p.second / 2)) % mod;
        }
      } else {
        ans = pw(ans, (p.second + 1));
      }
    }
  }
  cout << ans << endl;
  return 0;
}
