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;
long long int em(long long int x, long long int Mod, long long int n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return em((x % Mod * x % Mod) % Mod, Mod, n / 2) % Mod;
  else
    return (x % Mod * em((x % Mod * x % Mod) % Mod, Mod, (n - 1) / 2)) % Mod;
}
long long int cnt[200009];
int32_t main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, i, j, k, l;
  cin >> n;
  long long int ans = 1;
  long long int div = 1;
  set<long long int> s;
  for (long long int i = (long long int)0; i < (long long int)n; i++) {
    cin >> j;
    s.insert(j);
    cnt[j]++;
  }
  for (long long int y : s) {
    long long int tomul = em(y, 1000000007, ((cnt[y] + 1) * cnt[y]) / 2);
    ans = (em(ans, 1000000007, (cnt[y] + 1)) % 1000000007 *
           em(tomul, 1000000007, div) % 1000000007) %
          1000000007;
    div = (div * (cnt[y] + 1)) % (1000000007 - 1);
  }
  cout << ans % 1000000007 << "\n";
}
