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 powe(long long x, long long n, int long long mod) {
  long long ans = 1;
  while (n > 0) {
    if (n & 1) ans = (ans * x) % mod;
    n = n >> 1;
    x = (x * x) % mod;
  }
  return ans % mod;
}
int main() {
  long long m, i;
  long long flag = 0, a;
  long long d = 1;
  map<long long, long long>::iterator it;
  map<long long, long long> mp;
  cin >> m;
  for (i = 0; i < m; i++) {
    cin >> a;
    if (mp.find(a) == mp.end())
      mp.insert(make_pair(a, 1));
    else
      mp[a] = mp[a] + 1;
  }
  for (it = mp.begin(); it != mp.end(); it++) {
    a = it->second;
    if (a % 2 != 0 && flag == 0) {
      flag = 1;
      d = (((a + 1) / 2) * d) % (1000000007 - 1);
    } else
      d = ((a + 1) * d) % (1000000007 - 1);
  }
  long long p, ans = 1;
  for (it = mp.begin(); it != mp.end(); it++) {
    if (flag == 0)
      p = (((it->second / 2) * d) % (1000000007 - 1));
    else
      p = (it->second * d) % (1000000007 - 1);
    ans = (ans * powe(it->first, p, 1000000007)) % 1000000007;
  }
  cout << ans << endl;
  return 0;
}
