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 = 1000000007;
unordered_map<int, int> ja;
bool mark[maxn];
vector<long long> vec;
int m, a;
long long ans = 1, t = 1, num[maxn];
long long power(long long &a, int b, int m) {
  if (b == 0) return 1;
  long long hlp = power(a, b / 2, m);
  if (b & 1) return (((hlp * hlp) % m) * a) % m;
  return (hlp * hlp) % m;
}
inline void ghb() {
  for (int i = 2; i < maxn; i++) {
    for (int j = 2 * i; j < maxn; j += i) mark[j] = 1;
  }
  for (int i = 2; i < 200000; i++) {
    if (!mark[i]) {
      ja[i] = ((int(vec.size())));
      vec.push_back(i);
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ghb();
  cin >> m;
  for (int i = 0; i < m; i++) {
    cin >> a;
    num[ja[a]]++;
  }
  ans = power(vec[0], (num[0] * (num[0] + 1) / 2) % (mod - 1), mod);
  t = num[0] + 1;
  for (int i = 1; i < ((int(vec.size()))); i++) {
    ans = (((ans * power(ans, num[i], mod)) % mod) *
           power(vec[i],
                 (((num[i] * (num[i] + 1) / 2) % (mod - 1)) * t) % (mod - 1),
                 mod)) %
          mod;
    t = (t * (num[i] + 1)) % (mod - 1);
  }
  cout << ans;
  return 0;
}
