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 arr[200100];
long long mod = 1e9 + 7;
long long powe(long long a, long long b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  long long mid = powe(a, b / 2);
  if (b % 2 == 0) {
    return (mid * mid) % mod;
  } else {
    return (((mid * mid) % mod) * a) % mod;
  }
}
long long lhs[200100], rhs[200100];
int main() {
  long long i, j, k, n, m, t, cont;
  scanf("%lld", &n);
  for (i = 0; i < n; i++) {
    scanf("%lld", &m);
    arr[m]++;
  }
  long long prod = 1;
  lhs[0] = 1;
  for (i = 1; i < 200010; i++) {
    lhs[i] = (lhs[i - 1] * (arr[i] + 1)) % (mod - 1);
  }
  rhs[200001] = 1;
  for (i = 200000; i >= 0; i--) {
    rhs[i] = (rhs[i + 1] * (arr[i] + 1)) % (mod - 1);
  }
  long long ans = 1;
  for (i = 2; i < 200002; i++) {
    if (arr[i] != 0) {
      long long te;
      if (arr[i] % 2 == 0)
        te = ((arr[i] / 2) * (arr[i] + 1)) % (mod - 1);
      else
        te = ((arr[i]) * ((arr[i] + 1) / 2)) % (mod - 1);
      long long ju = (lhs[i - 1] * rhs[i + 1]) % (mod - 1);
      te = (te * ju) % (mod - 1);
      ans = (ans * powe(i, te)) % mod;
    }
  }
  printf("%lld\n", ans);
  return 0;
}
