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 int64_t N = 2e5 + 3;
const int64_t mod = 1e9 + 7;
int64_t dx[] = {-1, 0, 0, 1};
int64_t dy[] = {0, 1, -1, 0};
int64_t cnt[N];
int64_t expo(int64_t a, int64_t n) {
  int64_t ans = 1;
  while (n) {
    if (n % 2) ans = ans * a % mod;
    a = a * a % mod;
    n /= 2;
  }
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int64_t n;
  cin >> n;
  for (int64_t i = 1; i <= n; i++) {
    int64_t x;
    cin >> x;
    cnt[x]++;
  }
  int64_t ans = 1, d = 1;
  for (int64_t i = 2; i <= 2e5; i++) {
    int64_t m = cnt[i];
    int64_t w = expo(i, m * (m + 1) / 2);
    ans = expo(ans, m + 1) * expo(w, d) % mod;
    d = d * (m + 1) % (mod - 1);
  }
  cout << ans;
}
