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 pow(long long x, long long n) {
  long long res = 1;
  for (; n; n >>= 1) {
    if (n & 1) {
      res *= x;
      res %= 1000000007LL;
    }
    x *= x;
    x %= 1000000007LL;
  }
  return res;
}
bool is_odd(long long x) { return x % 2 == 1; }
int main(int argc, char const *argv[]) {
  long long m;
  cin >> m;
  vector<long long> p(200001, 0);
  for (int i = 0; i < m; ++i) {
    long long x;
    cin >> x;
    ++p[x];
  }
  if (any_of(p.begin(), p.end(), is_odd)) {
    long long n = 1;
    long long deg = 1;
    bool is_divided_by_2 = false;
    for (long long i = 1; i < 200000; ++i) {
      n *= pow(i, p[i]);
      n %= 1000000007LL;
      if (!is_divided_by_2 && p[i] % 2 == 1) {
        deg *= (p[i] + 1) / 2;
        deg %= 1000000007LL - 1;
        is_divided_by_2 = true;
      } else
        deg *= p[i] + 1;
      deg %= 1000000007LL - 1;
    }
    cout << pow(n, deg) << endl;
  } else {
    long long n = 1;
    long long deg = 1;
    for (long long i = 1; i < 200000; ++i) {
      n *= pow(i, p[i] / 2);
      n %= 1000000007LL;
      deg *= p[i] + 1;
      deg %= 1000000007LL - 1;
    }
    cout << pow(n, deg) << endl;
  }
  return 0;
}
