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 binexp(long long a, long long b, long long m) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % m;
    a = (a * a) % m;
    b >>= 1;
  }
  return res;
}
int cnt[2000005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  set<int> st;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    st.insert(x);
    cnt[x]++;
  }
  vector<int> arr(st.begin(), st.end());
  long long a = 1;
  int f = 0;
  for (int i = 0; i < arr.size(); i++) {
    if ((cnt[arr[i]] & 1) && !f) {
      a = (a * ((1LL + cnt[arr[i]]) / 2)) % (1000000007LL - 1);
      f = 1;
    } else {
      a = (a * (1LL + cnt[arr[i]])) % (1000000007LL - 1);
    }
  }
  long long ans = 1;
  for (int i = 0; i < arr.size(); i++) {
    long long t;
    if (f == 0)
      t = (a * (cnt[arr[i]] / 2)) % (1000000007LL - 1);
    else
      t = (a * cnt[arr[i]]) % (1000000007LL - 1);
    ans = (ans * binexp(arr[i], t, 1000000007LL)) % 1000000007LL;
  }
  cout << ans << "\n";
}
