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 int N = 2e5 + 5, OO = 0x3f3f3f3f, MOD = 1000000007;
const double EPS = 0.000000001;
int m, x;
long long p = 1, ans = 1;
unordered_map<int, int> freq;
long long power(int b, int p) {
  if (!p) return 1;
  long long sub = power(b, p >> 1) % MOD;
  return (p & 1 ? (((sub * sub) % MOD * b) % MOD) : (sub * sub) % MOD);
}
int main() {
  cin.sync_with_stdio(0);
  cin >> m;
  for (int i = 0; i < m; ++i) {
    cin >> x;
    ++freq[x];
  }
  int MODP = MOD - 1;
  bool f = 1;
  for (pair<int, int> factor : freq) {
    int e = factor.second + 1;
    if (f && e % 2 == 0) f = 0, e /= 2;
    p = (p * e) % MODP;
  }
  for (pair<int, int> factor : freq) {
    int e = factor.second;
    if (f) e /= 2;
    long long term = power(factor.first, (p * e) % MODP) % MOD;
    ans = (ans * term) % MOD;
  }
  printf("%lld\n", ans);
  return 0;
}
