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 int Maxn = 200 * 1000 + 10, Mod = 1000 * 1000 * 1000 + 7;
long long m, arr[Maxn], num = 1, ans = 1;
vector<pair<long long, long long> > vec;
long long powe(long long a, long long b) {
  long long ans1 = 1;
  for (int i = 63; i >= 0; i--) {
    ans1 *= ans1;
    ans1 %= Mod;
    if ((b >> i) & 1) ans1 *= a;
    ans1 %= Mod;
  }
  return ans1;
}
int main() {
  scanf("%I64d", &m);
  for (int i = 0; i < m; i++) scanf("%I64d", &arr[i]);
  sort(arr, arr + m);
  vec.push_back(make_pair(arr[0], 1));
  for (int i = 1; i < m; i++) {
    if (arr[i] == arr[i - 1])
      vec[(int)vec.size() - 1].second++;
    else
      vec.push_back(make_pair(arr[i], 1));
  }
  for (int i = 0; i < (int)vec.size(); i++) {
    ans *= powe(ans, vec[i].second);
    ans %= Mod;
    ans *=
        powe(powe(vec[i].first, vec[i].second * (vec[i].second + 1) / 2), num);
    ans %= Mod;
    num *= (vec[i].second + 1);
    num %= (Mod - 1);
  }
  printf("%I64d\n", ans);
  return 0;
}
