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 mod = (long long)(1e9 + 7);
const int maxn = 200005;
int m;
long long p[maxn];
int cnt[maxn];
long long power(long long a, long long pp) {
  long long res = 1;
  while (pp) {
    if (pp & 1) res = res * a;
    res %= mod;
    a = a * a;
    a %= mod;
    pp >>= 1;
  }
  return res;
}
int main() {
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%I64d", &p[i]);
    cnt[p[i]]++;
  }
  long long sum = 1;
  for (int i = 2; i <= 200000; i++) {
    if (cnt[i]) {
      sum *= (cnt[i] + 1);
      sum %= ((mod - 1) * 2);
    }
  }
  long long ans = 1;
  for (int i = 2; i <= 200000; i++) {
    if (cnt[i]) {
      long long tmp = (sum * cnt[i] / 2) % (mod - 1);
      ans *= power(i, tmp);
      ans %= mod;
    }
  }
  printf("%I64d\n", ans % mod);
  return 0;
}
