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 int lim = 100001;
long long int powe(long long int a, long long int b) {
  long long int ans = 1;
  while (b) {
    if (b & 1) {
      ans = ((ans % 1000000007) * (a % 1000000007)) % 1000000007;
    }
    a = ((a % 1000000007) * (a % 1000000007)) % 1000000007;
    b = b >> 1;
  }
  return ans;
}
bool chk(string s, long long int i, vector<long long int> color) {
  char cur = s[i];
  long long int j = i - 1;
  while (j >= 0) {
    if (s[j] > cur) {
      if (color[j] != -1) {
        if (color[i] == color[j]) {
          return 0;
        }
      }
    }
    j--;
  }
  return 1;
}
void solve() {
  long long int n;
  cin >> n;
  map<long long int, long long int> mp;
  for (int i = 0; i < n; i++) {
    long long int temp;
    cin >> temp;
    mp[temp]++;
  }
  long long int ans = 1;
  long long int prev = 1;
  for (auto i : mp) {
    long long int freq = i.second;
    long long int tempans = powe(i.first, (freq * (freq + 1)) / 2);
    ans = (powe(ans, freq + 1) * powe(tempans, prev)) % 1000000007;
    prev = (prev * (freq + 1)) % (1000000007 - 1);
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t = 1;
  while (t > 0) {
    string s1, s2;
    solve();
    t--;
  }
}
