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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const long long MOD = (long long)1e9 + 7;
long long binPow(long long a, long long q, long long MOD) {
  a %= MOD;
  if (q == 0) return 1;
  return ((q % 2 == 1 ? a : 1) * binPow(a * a, q / 2, MOD)) % MOD;
}
int main() {
  int n;
  scanf("%d", &n);
  map<int, int> cnt;
  while (n--) {
    int x;
    scanf("%d", &x);
    cnt[x]++;
  }
  long long d = 1;
  long long ans = 1;
  for (auto x : cnt) {
    long long cnt = x.second;
    long long p = x.first;
    long long fp = binPow(p, (cnt + 1) * cnt / 2, MOD);
    ans = binPow(ans, (cnt + 1), MOD) * binPow(fp, d, MOD) % MOD;
    d = d * (x.second + 1) % (MOD - 1);
  }
  cout << ans << endl;
  return 0;
}
