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 Mod = 1000000000 + 7;
long long power(int x, long long p) {
  long long ret = 1;
  while (p) {
    if (p & 1) ret = 1LL * ret * x % Mod;
    x = 1LL * x * x % Mod;
    p >>= 1;
  }
  return ret;
}
int m;
const int MAXN = 200000;
int a[MAXN + 10];
void init() {
  cin >> m;
  long long ans = 1, q = 1;
  for (int i = 1; i <= m; i++) scanf("%d", &a[i]);
  sort(a + 1, a + m + 1);
  for (int i = 1; i <= m; i++) {
    int now = a[i], j = i;
    long long p = 1;
    while (j < m && a[j + 1] == a[i]) {
      p++;
      j++;
    }
    ans = 1LL * power(ans, p + 1) *
          power(a[i], (1LL * p * (p + 1) >> 1) % (Mod - 1) * q % (Mod - 1)) %
          Mod;
    q = q * (p + 1) % (Mod - 1);
    i = j;
  }
  cout << ans;
  return;
}
int main() {
  init();
  return 0;
}
