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 lt = 2e5;
long long cnt[200002];
long long power(long long A, long long B) {
  A %= 1000000007;
  long long res = 1;
  while (B > 0LL) {
    if (B & 1LL) res = (res * A) % 1000000007;
    A = (A * A) % 1000000007;
    B >>= 1LL;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL), cout.tie(NULL);
  long long m, i, inp;
  cin >> m;
  for (i = 1; i <= m; i++) {
    cin >> inp;
    cnt[inp]++;
  }
  long long f = 0;
  for (i = 1; i <= lt; i++) {
    if (cnt[i] == 0) continue;
    if (cnt[i] % 2 == 1) {
      f = 1;
      break;
    }
  }
  long long pr = 1;
  if (f == 1) {
    for (i = 1; i <= lt; i++) {
      pr = (pr * power(i, cnt[i])) % 1000000007;
      cnt[i]++;
    }
    for (i = 1; i <= lt; i++) {
      if (cnt[i] % 2 == 0) {
        cnt[i] = cnt[i] / 2;
        break;
      }
    }
    for (i = 1; i <= lt; i++) {
      pr = power(pr, cnt[i]);
    }
  } else {
    for (i = 1; i <= lt; i++) {
      pr = (pr * power(i, cnt[i] / 2)) % 1000000007;
      cnt[i]++;
    }
    for (i = 1; i <= lt; i++) {
      pr = power(pr, cnt[i]);
    }
  }
  cout << pr << endl;
  return 0;
}
