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;
long long int bigmod(long long int a, long long int b, long long int c) {
  long long int x = 1;
  long long int y = a % c;
  while (b > 0) {
    if (b % 2 == 1) x = (x * y) % c;
    y = (y * y) % c;
    b /= 2;
  }
  return x % c;
}
int main() {
  long long int n, i, k, ans, den, p, prod = 1, srt = 1, power = 1, j, flag = 0,
                                      d = 0;
  map<long long int, long long int> mp;
  map<long long int, long long int>::iterator it;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> k;
    prod = (prod * k) % 1000000007;
    mp[k]++;
  }
  for (it = mp.begin(); it != mp.end(); it++) {
    p = (*it).second;
    k = (*it).first;
    power = (power * (p + 1)) % (2 * 1000000007 - 2);
    if (p % 2 == 1) flag = 1;
    for (j = 1; j <= p / 2; j++) {
      srt = (srt * k) % 1000000007;
    }
  }
  power = (power / 2);
  ans = bigmod(prod, power, 1000000007);
  if (flag == 0) {
    ans = (ans * srt) % 1000000007;
  }
  cout << ans << "\n";
  return 0;
}
