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 M = 2 * (1000000007 - 1);
int n, cnt[200005], a[200005];
int pmod(int a, int n) {
  int ans = 1;
  while (n) {
    if (n & 1) ans = (1ll * ans * a) % 1000000007;
    a = (1ll * a * a) % 1000000007;
    n >>= 1;
  }
  return ans;
}
int main() {
  cin >> n;
  int ans = 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    ans = (1ll * ans * a[i]) % 1000000007;
    cnt[a[i]]++;
  }
  int x = 1;
  bool ok = 1;
  for (int i = 1; i < 200005; i++) {
    x = (1ll * x * (cnt[i] + 1)) % M;
    if (cnt[i] & 1) ok = 0;
  }
  x /= 2;
  ans = pmod(ans, x);
  if (ok)
    for (int i = 1; i < 200005; i++)
      ans = (1ll * ans * pmod(i, cnt[i] / 2)) % 1000000007;
  cout << ans;
  return 0;
}
