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 int maxn = 200000 + 10;
const long long mod = 1e9 + 7;
int cnt[maxn];
long long fun(long long p, long long k) {
  long long res = 1;
  while (k) {
    if ((k & 1) == 1) res = res * p % mod;
    p = p * p % mod;
    k >>= 1;
  }
  return res % mod;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int k;
    scanf("%d", &k);
    cnt[k]++;
  }
  long long M = 1, ans = 1;
  for (int i = 1; i < maxn; i++) {
    M = M * (cnt[i] + 1) % (2 * mod - 2);
  }
  for (int i = 1; i < maxn; i++) {
    long long tmp = M * cnt[i] % (2 * mod - 2) / 2;
    ans = ans * fun(i, tmp) % mod;
  }
  printf("%I64d\n", ans);
  return 0;
}
