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 = 1e5, M = 3e5, OO = 0x3f3f3f3f, MOD = 1e9 + 7;
int freq[200000], m, a, mx, mn = OO;
long long fastpower(long long x, long long p) {
  long long res = 1;
  while (p) {
    if (p & 1) res = (res * x) % (MOD);
    x = (x * x) % (MOD);
    p >>= 1;
  }
  return res;
}
int main() {
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d", &a);
    freq[a]++;
    mx = max(mx, a);
    mn = min(mn, a);
  }
  long long d = 1;
  bool flag = false;
  for (int i = mn; i <= mx; i++) {
    if (freq[i]) {
      if (!flag && (freq[i] & 1)) {
        flag = true;
        d = (d * ((freq[i] + 1) / 2)) % (MOD - 1);
      } else
        d = (d * (freq[i] + 1)) % (MOD - 1);
    }
  }
  long long f = 1;
  for (int i = mn; i <= mx; i++) {
    if (freq[i]) {
      if (!flag) {
        freq[i] /= 2;
      }
      f = (f * fastpower(i, d * freq[i])) % MOD;
    }
  }
  printf("%I64d", f);
}
