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 my_pow(long long _x, long long _n) {
  long long ret = 1;
  while (0 < _n) {
    if ((_n % 2) == 0) {
      _x = (_x * _x) % 1000000007LL;
      _n >>= 1;
    } else {
      ret = (ret * _x) % 1000000007LL;
      --_n;
    }
  }
  return ret;
}
map<long long, long long> fm;
int main() {
  long long m, n = 1;
  cin >> m;
  long long p;
  for (long long i = 0; i < m; i++) {
    cin >> p;
    if (fm.find(p) == fm.end()) {
      fm[p] = 1;
    } else {
      fm[p]++;
    }
  }
  long long ans = 1;
  bool issq = true;
  long long tp = -1;
  for (auto pp : fm) {
    if (issq && pp.second % 2 == 1) {
      issq = false;
      tp = pp.first;
    }
  }
  if (issq) {
    for (auto pp : fm) {
      n = (n * my_pow(pp.first, pp.second / 2)) % 1000000007LL;
    }
    ans = n;
    for (auto pp : fm) {
      ans = my_pow(ans, pp.second + 1) % 1000000007LL;
    }
  } else {
    for (auto pp : fm) {
      n = (n * my_pow(pp.first, pp.second)) % 1000000007LL;
    }
    ans = n;
    for (auto pp : fm) {
      if (pp.first == tp) {
        if ((pp.second + 1) / 2 > 1) {
          ans = my_pow(ans, (pp.second + 1) / 2) % 1000000007LL;
        }
      } else {
        ans = my_pow(ans, pp.second + 1) % 1000000007LL;
      }
    }
  }
  cout << ans << endl;
}
