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 N = 200000;
const int MOD = (int)1e9 + 7;
int f[N + 9];
int pf[N + 9];
int sf[N + 9];
int binpow(int a, int n) {
  int r = 1;
  while (n) {
    if (n & 1) {
      r = (r * 1LL * a) % MOD;
    }
    a = (a * 1LL * a) % MOD;
    n >>= 1;
  }
  return r;
}
int sum(int n) { return (n * 1LL * (n + 1)) / 2 % (MOD - 1); }
int main() {
  int m;
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    int p;
    scanf("%d", &p);
    f[p]++;
  }
  pf[0] = 1;
  for (int i = 1; i <= N; i++) {
    pf[i] = pf[i - 1] * 1LL * (f[i] + 1) % (MOD - 1);
  }
  sf[N + 1] = 1;
  for (int i = N; i >= 1; i--) {
    sf[i] = sf[i + 1] * 1LL * (f[i] + 1) % (MOD - 1);
  }
  int ans = 1;
  for (int i = 1; i <= N; i++) {
    ans = ans * 1LL *
          binpow(i, sum(f[i]) * 1LL * pf[i - 1] % (MOD - 1) * 1LL * sf[i + 1] %
                        (MOD - 1)) %
          MOD;
  }
  printf("%d\n", ans);
  return 0;
}
