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 bp(long long a, long long n) {
  if (n == 0) return 1;
  if (n == 1) return a;
  long long ans = 1;
  if (n & 1) {
    ans = (ans * a) % ((long long)1e9 + 7);
  }
  return (ans * bp((a * a) % ((long long)1e9 + 7), n / 2)) %
         ((long long)1e9 + 7);
}
int main() {
  long long n, s = 1;
  cin >> n;
  long long a[n];
  map<int, int> mp;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    mp[a[i]]++;
  }
  bool test = false;
  for (int i = 0; i < 200200; i++) {
    test = test || (mp[i] & 1);
  }
  if (test) {
    test = false;
    for (int i = 0; i < n; i++) s = (s * a[i]) % ((long long)1e9 + 7);
    for (int i = 0; i < 200200; i++) {
      if ((mp[i] & 1) && !test) {
        s = (bp(s, (mp[i] + 1) / 2)) % ((long long)1e9 + 7);
        test = true;
      } else {
        s = (bp(s, (mp[i] + 1))) % ((long long)1e9 + 7);
      }
    }
  } else {
    for (int i = 0; i < 200200; i++) {
      for (int j = 0; j < mp[i] / 2; j++) {
        s = (s * i) % ((long long)1e9 + 7);
      }
    }
    for (int i = 0; i < 200200; i++) {
      s = (bp(s, mp[i] + 1)) % ((long long)1e9 + 7);
    }
  }
  cout << s;
  return 0;
}
