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;
map<long long, long long> primeFactor;
long long m, x;
long long d = 1, answer = 1, cnt, p, fp;
const long long MOD = (long long)1e9 + 7;
long long binPow(long long a, long long q, long long MOD) {
  a %= MOD;
  if (q == 0) return 1;
  return ((q % 2 == 1 ? a : 1) * binPow(a * a, q / 2, MOD)) % MOD;
}
int main() {
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d", &x);
    primeFactor[x]++;
  }
  for (auto& [p, cnt] : primeFactor) {
    fp = binPow(p, (cnt + 1) * cnt / 2, MOD);
    answer = binPow(answer, (cnt + 1), MOD) * binPow(fp, d, MOD) % MOD;
    d = d * (cnt + 1) % (MOD - 1);
  }
  printf("%lld", answer);
  return 0;
}
