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 mod = 1000000007;
const long long mod2 = 2000000012;
long long power(long long a, long long m) {
  long long ans = 1;
  while (m) {
    if (m & 1) ans = ans * a % mod;
    a = a * a % mod;
    m >>= 1;
  }
  return ans;
}
int main() {
  map<int, int> p;
  map<int, int>::iterator f;
  int m, q;
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d", &q);
    f = p.find(q);
    if (f != p.end())
      (f->second)++;
    else
      p.insert(make_pair(q, 1));
  }
  long long all = 1;
  for (f = p.begin(); f != p.end(); f++) all = all * ((f->second) + 1) % mod2;
  long long ans = 1;
  for (f = p.begin(); f != p.end(); f++) {
    long long exp = (f->second) * all % mod2;
    exp /= 2;
    ans = ans * power((f->first), exp) % mod;
  }
  printf("%lld", ans);
}
