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 = 1000000007LL;
map<long long, long long> mp;
long long fp(long long a, long long k) {
  long long res = 1;
  while (k) {
    if (k & 1) res = res * a % Mod;
    a = a * a % Mod, k >>= 1;
  }
  return res;
}
int main() {
  long long m, p, s = 1, res = 1;
  scanf("%I64d", &m);
  while (m--) mp[scanf("%I64d", &p), p]++;
  for (auto itr = mp.begin(); itr != mp.end(); itr++)
    (s *= (itr->second + 1)) %= 2 * Mod - 2;
  for (auto itr = mp.begin(); itr != mp.end(); itr++)
    (res *= fp(itr->first, itr->second * s % (2 * Mod - 2) / 2)) %= Mod;
  return 0 * printf("%I64d\n", res);
}
