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 = 1E9 + 7;
long long mod_pow(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = (ans * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return ans % mod;
}
long long d(pair<long long, long long> p) { return p.second + 1; }
long long f(pair<long long, long long> p) {
  return mod_pow(p.first, (p.second * (p.second + 1) / 2) % (mod - 1));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int m;
  cin >> m;
  map<long long, long long> p;
  for (int i = 0; i < m; i++) {
    long long aux;
    cin >> aux;
    p[aux]++;
  }
  long long ans = f(*p.begin()), divs = d(*p.begin());
  for (auto &x : p) {
    if (x == *p.begin()) continue;
    ans = mod_pow(ans, d(x)) * mod_pow(f(x), divs);
    ans %= mod;
    divs *= d(x);
    divs %= mod - 1;
  }
  cout << ans % mod << endl;
  return 0;
}
