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;
map<int, long long> help;
map<int, long long>::iterator it;
map<int, int> done;
int modulo(int a, long long b, int c) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y) % c;
    }
    y = (y * y) % c;
    b /= 2;
  }
  return x % c;
}
int main() {
  int n;
  scanf("%d", &n);
  int a;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a);
    if (help.find(a) == help.end()) {
      help[a] = 1;
    } else {
      help[a]++;
    }
  }
  long long prod = 1;
  long long div = 1;
  int val = -1;
  for (it = help.begin(); it != help.end(); it++) {
    if ((it->second) % 2 == 1) {
      val = it->first;
      break;
    }
  }
  if (val == -1) {
    for (it = help.begin(); it != help.end(); it++) {
      div = (div * (it->second + 1)) % (1000000007 - 1);
    }
  } else {
    for (it = help.begin(); it != help.end(); it++) {
      if (it->first != val) {
        div = (div * (it->second + 1)) % (1000000007 - 1);
      } else {
        div = (div * ((it->second + 1) / 2)) % (1000000007 - 1);
      }
    }
  }
  for (it = help.begin(); it != help.end(); it++) {
    if (val != -1) {
      long long temp = modulo(
          it->first, (((it->second) * div)) % (1000000007 - 1), 1000000007);
      prod = (prod * temp) % 1000000007;
    } else {
      long long temp =
          modulo(it->first, ((((it->second) / 2) * div)) % (1000000007 - 1),
                 1000000007);
      prod = (prod * temp) % 1000000007;
    }
  }
  printf("%lld", prod);
}
