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 long long mod = 1000000007LL;
long long mod_pow(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n % 2) {
      res = (res * x) % mod;
    }
    x = (x * x) % mod;
    n /= 2;
  }
  return res;
}
int main() {
  int m;
  cin >> m;
  map<long long, long long> cnt;
  for (int i = 0; i < m; i++) {
    long long p;
    cin >> p;
    cnt[p]++;
  }
  vector<pair<long long, long long>> deg;
  for (auto it = cnt.begin(); it != cnt.end(); it++) {
    deg.push_back(*it);
  }
  int M = deg.size();
  vector<long long> L(M), R(M);
  L[0] = deg[0].second + 1;
  R[M - 1] = deg[M - 1].second + 1;
  for (int i = 1; i < M; i++)
    L[i] = (L[i - 1] * (deg[i].second + 1)) % (mod - 1);
  for (int i = M - 2; i >= 0; i--)
    R[i] = (R[i + 1] * (deg[i].second + 1)) % (mod - 1);
  long long res = 1;
  for (int i = 0; i < M; i++) {
    long long num = (deg[i].second * (deg[i].second + 1) / 2) % (mod - 1);
    if (i != 0) num = (num * L[i - 1]) % (mod - 1);
    if (i != M - 1) num = (num * R[i + 1]) % (mod - 1);
    res = (res * mod_pow(deg[i].first, num)) % mod;
  }
  cout << res << endl;
  return 0;
}
