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 pw(int a, int p, int md) {
  if (p == 0) return 1;
  long long r = pw(a, p / 2, md);
  r *= r;
  r %= md;
  if (p % 2 == 1) {
    return (a * r) % md;
  }
  return r;
}
vector<int> pr;
vector<int> cnt;
int p[200005];
int main() {
  int m;
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d", p + i);
  }
  sort(p, p + m);
  for (int i = 0; i < m; i++) {
    if (i == 0 || p[i] != p[i - 1]) {
      pr.push_back(p[i]);
      cnt.push_back(1);
    } else {
      cnt[cnt.size() - 1]++;
    }
  }
  const int mod = 1000000007;
  const int mod2 = 1000000006;
  long long divs[200005];
  for (int i = 0; i < pr.size(); i++) {
    if (i)
      divs[i] = divs[i - 1];
    else
      divs[i] = 1;
    divs[i] *= (cnt[i] + 1);
    divs[i] %= mod2;
  }
  long long divs2[200005];
  for (int i = pr.size() - 1; i >= 0; i--) {
    if (i + 1 < pr.size())
      divs2[i] = divs2[i + 1];
    else
      divs2[i] = 1;
    divs2[i] *= (cnt[i] + 1);
    divs2[i] %= mod2;
  }
  long long res = 1;
  for (int i = 0; i < pr.size(); i++) {
    long long other = 1;
    if (i) other *= divs[i - 1];
    if (i + 1 < pr.size()) {
      other *= divs2[i + 1];
    }
    other %= mod2;
    long long a = 1;
    for (int j = 0; j < cnt[i]; j++) {
      a *= pr[i];
      a %= mod;
      res *= pw(a, other, mod);
      res %= mod;
    }
  }
  cout << res << endl;
}
