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;
vector<int> v;
long long pow1(long long a, long long N) {
  if (N == 0) return 1;
  if (N == 1) return a;
  long long temp = pow1(a, N >> 1);
  temp *= temp;
  temp %= 1000000007;
  if (N % 2 == 1) temp *= a;
  temp %= 1000000007;
  return (temp);
}
map<int, int> mp;
int main() {
  int N;
  int temp;
  long long ans = 1;
  scanf("%d", &N);
  while (N--) {
    scanf("%d", &temp);
    mp[temp]++;
  }
  int M = (1000000007 - 1) * 2;
  long long mul = 1;
  for (map<int, int>::iterator it = mp.begin(); it != mp.end(); it++) {
    mul *= (it->second + 1);
    mul %= M;
  }
  for (map<int, int>::iterator it = mp.begin(); it != mp.end(); it++) {
    long long temp1 = mul * (long long)(it->second);
    temp1 %= M;
    ans *= pow1(it->first, temp1 / 2);
    ans %= 1000000007;
  }
  cout << ans << endl;
  return 0;
}
