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 = 210000;
const long long MOD = 1e9 + 7;
long long a[N], p[N], r[N], n, dx;
long long Pow(long long x, long long y, long long mod) {
  long long ret = 1;
  while (y) {
    if (y & 1) ret = ret * x % mod;
    x = x * x % mod;
    y = y >> 1;
  }
  return ret;
}
long long f() {
  int cur = 0;
  sort(a, a + n);
  p[0] = a[0];
  r[0] = 1;
  for (int i = 1; i < n; i++) {
    if (a[i] == p[cur])
      r[cur]++;
    else {
      p[++cur] = a[i];
      r[cur] = 1;
    }
  }
  long long ret = 1;
  long long ans = 1;
  long long fp = 1;
  for (int i = 0; i <= cur; i++) {
    fp = Pow(p[i], (r[i] + 1) * r[i] / 2, MOD);
    ans = Pow(ans, r[i] + 1, MOD) * Pow(fp, ret, MOD) % MOD;
    ret = ret * (r[i] + 1) % (MOD - 1);
  }
  return ans;
}
int main() {
  cin >> n;
  long long m = 1;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  cout << f() << endl;
  return 0;
}
