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 n;
long long mod = 1e9 + 7, a[300000];
long long binpow(long long x, long long y) {
  if (y == 0) return 1;
  if (y % 2) return x * binpow(x, y - 1) % mod;
  long long t = binpow(x, y / 2);
  return t * t % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    a[x]++;
  }
  long long s = 1, ans = 1;
  int k = 1;
  for (long long i = 1; i <= 2e5; i++) {
    if (k == 1 && a[i] % 2) {
      s = s = s * ((a[i] + 1) / 2) % (mod - 1);
      k = 0;
    } else
      s = s * (a[i] + 1) % (mod - 1);
  }
  for (long long i = 1; i <= 2e5; i++)
    if (a[i]) {
      long long t;
      if (k == 1)
        t = a[i] * s / 2;
      else
        t = a[i] * s;
      ans = (ans * binpow(i, t % (mod - 1))) % mod;
    }
  cout << ans << endl;
}
