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 powermodm(long long x, long long n, long long M) {
  long long result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % M;
    x = (x * x) % M;
    n = n / 2;
  }
  return result;
}
map<long long, long long> m;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    m[a[i]]++;
  }
  long long sz = m.size();
  long long ans = 1;
  long long cnt = 1, c = 1, mod2 = 1;
  long long flag = 0;
  for (auto it = m.begin(); it != m.end(); it++) {
    long long x = it->second;
    long long f = powermodm(it->first, (x * (x + 1)) / 2, 1000000007);
    long long p = powermodm(ans, x + 1, 1000000007);
    ans = (p * (powermodm(f, cnt, 1000000007))) % 1000000007;
    cnt = (cnt * (x + 1)) % (1000000007 - 1);
  }
  cout << ans;
}
