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 = 222222;
const int mo = 1e9 + 7;
int cnt[N];
int pre[N], suf[N];
int m;
int get(int x, int k) {
  int ret = 1;
  while (k) {
    if (k & 1) {
      ret = 1ll * ret * x % mo;
    }
    k >>= 1;
    x = 1ll * x * x % mo;
  }
  return ret;
}
int main() {
  scanf("%d", &m);
  for (int i = 0; i < m; ++i) {
    int x;
    scanf("%d", &x);
    cnt[x]++;
  }
  pre[0] = suf[N] = 1;
  for (int i = 1; i < N; ++i) {
    pre[i] = 1ll * pre[i - 1] * (cnt[i] + 1) % (mo - 1);
  }
  for (int i = N - 1; i > 0; --i) {
    suf[i] = 1ll * suf[i + 1] * (cnt[i] + 1) % (mo - 1);
  }
  int ans = 1;
  for (int i = 1; i < N; ++i) {
    if (cnt[i] == 0) continue;
    int curBase = get(i, 1ll * pre[i - 1] * suf[i + 1] % (mo - 1));
    int cur = 1;
    for (int j = 0; j < cnt[i]; ++j) {
      cur = 1ll * cur * curBase % mo;
      ans = 1ll * ans * cur % mo;
    }
  }
  printf("%d\n", ans);
  return 0;
}
