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>
const int mod = 1e9 + 7;
long long pow_mod(long long a, long long k, int mod) {
  long long s = 1;
  while (k) {
    if (k & 1) s = s * a % mod;
    a = a * a % mod;
    k >>= 1;
  }
  return s;
}
int p[200001];
int main() {
  int m;
  scanf("%d", &m);
  for (int i = 0; i < m; ++i) {
    int t = 0;
    scanf("%d", &t);
    p[t]++;
  }
  long long prod = 1;
  for (int i = 0; i <= 200000; ++i) {
    prod = prod * (1 + p[i]) % ((mod - 1) * 2);
  }
  long long ans = 1;
  for (int i = 0; i <= 200000; ++i) {
    if (p[i]) {
      long long cnt = prod * p[i] / 2;
      ans = ans * pow_mod(i, cnt, mod) % mod;
    }
  }
  printf("%I64d\n", ans);
  return 0;
}
