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 long long M = 1e9 + 7, re = 500000004;
const int MAX = 2e5 + 1;
long long d[MAX], n;
inline long long f(long long a, long long q, long long MOD = M) {
  a %= MOD;
  if (q == 0) return 1;
  return ((q % 2 == 1 ? a : 1) * f(a * a, q / 2, MOD)) % MOD;
}
inline long long mul(long long x, long long y) { return x * y % M; }
inline long long sum(long long x, long long y) { return (x + y) % M; }
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    int j;
    cin >> j;
    ++d[j];
  }
  long long ans = 1;
  long long cur = 1;
  for (int i = 2; i < MAX; i++) {
    long long p1 = f(f(i, d[i] * (d[i] + 1) / 2), cur);
    long long p2 = f(ans, d[i] + 1);
    ans = mul(p1, p2);
    cur = cur * (d[i] + 1) % (M - 1);
  }
  cout << ans << '\n';
  return 0;
}
