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 int power(long long int a, long long int b) {
  long long int ans = 1;
  long long int t = a;
  while (b) {
    if (b % 2) {
      ans = (ans * t) % 1000000007;
    }
    t = (t * t) % 1000000007;
    b /= 2;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  long long int a[n];
  int f[200001];
  memset(f, 0, sizeof f);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    ++f[a[i]];
  }
  long long int factor = 1;
  for (int fr : f) {
    factor = (factor * (fr + 1)) % 1000000007;
  }
  long long int pre[200001], suf[200001];
  pre[0] = 1, suf[200000] = 1;
  for (int i = 1; i < 200001; i++) {
    pre[i] = (pre[i - 1] * (f[i] + 1)) % (1000000007 - 1);
  }
  for (int i = 200000 - 1; i >= 0; i--) {
    suf[i] = ((f[i] + 1) * suf[i + 1]) % (1000000007 - 1);
  }
  long long int ans = 1;
  for (long long int p = 1; p < 200001; p++) {
    long long int oth = (pre[p - 1] * suf[p + 1]) % (1000000007 - 1);
    long long int h = 1;
    for (int i = 0; i < f[p]; i++) {
      h = (h * p) % 1000000007;
      ans = (ans * power(h, oth)) % 1000000007;
    }
  }
  cout << ans;
}
