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;
using ll = long long;
using ld = long double;
const int MOD = 1e9 + 7;
const int N = 2e5;
int fastexpo(int a, int b) {
  int ret = 1;
  while (b) {
    if (b & 1) {
      ret = 1LL * ret * a % MOD;
    }
    a = 1LL * a * a % MOD;
    b >>= 1;
  }
  return ret;
}
int cnt[N + 5], pre[N + 5], suf[N + 5];
int m;
void solve() {
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int x;
    cin >> x;
    cnt[x]++;
  }
  pre[0] = 1, suf[N + 1] = 1;
  for (int i = 1; i <= N; i++) {
    pre[i] = 1LL * pre[i - 1] * (cnt[i] + 1) % (MOD - 1);
  }
  for (int i = N; i >= 1; i--) {
    suf[i] = 1LL * suf[i + 1] * (cnt[i] + 1) % (MOD - 1);
  }
  int ans = 1;
  for (int i = 1; i <= N; i++) {
    if (cnt[i] == 0) continue;
    int temp = 1LL * pre[i - 1] * suf[i + 1] % (MOD - 1);
    int cur = (1LL * cnt[i] * (cnt[i] + 1) / 2) % (MOD - 1);
    temp = 1LL * temp * cur % (MOD - 1);
    ans = 1LL * ans * fastexpo(i, temp) % MOD;
  }
  if (ans < 0) ans += MOD;
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int TC = 1;
  for (int i = 1; i <= TC; i++) {
    solve();
  }
  return 0;
}
