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 LIM = 2e+5 + 100;
const long long MOD = 1e+9 + 7;
int cnt[LIM];
void mod_mult(long long& x, long long y, long long mod = MOD) {
  x *= y;
  x %= mod;
}
long long bin_pow(long long x, long long p, long long mod = MOD) {
  if (p == 0) return 1;
  if (p == 1) return x;
  long long t = bin_pow(x, p >> 1);
  mod_mult(t, t, mod);
  if (p & 1) mod_mult(t, x, mod);
  return t;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    int tmp;
    cin >> tmp;
    ++cnt[tmp];
  }
  long long x = 1;
  long long pw = 1;
  long long ans = 1;
  for (int i = 1; i < LIM; ++i) {
    if (cnt[i] == 0) continue;
    long long tmp = bin_pow(i, cnt[i]);
    long long f_tmp = 1;
    for (int j = 1; j <= cnt[i]; ++j) mod_mult(f_tmp, bin_pow(i, j));
    long long first = bin_pow(ans, cnt[i] + 1);
    long long second = bin_pow(f_tmp, pw);
    ans = first;
    mod_mult(ans, second);
    mod_mult(pw, cnt[i] + 1, MOD - 1);
    mod_mult(x, tmp);
  }
  cout << ans << endl;
  return 0;
}
