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 = 200005, phi = (long long)(1e9 + 7) - 1;
long long cnt[N], pref[N], suff[N];
long long binpow(long long n, long long a, long long p) {
  long long res = 1ll % p;
  while (a) {
    if (a & 1) {
      res *= n;
      res %= p;
    }
    n *= n;
    n %= p;
    a >>= 1;
  }
  return res;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    cnt[x]++;
  }
  pref[0] = 1;
  for (int i = 1; i < N; i++) pref[i] = (pref[i - 1] * (cnt[i] + 1)) % phi;
  suff[N - 1] = 1;
  for (int i = N - 2; i > 0; i--) suff[i] = (suff[i + 1] * (cnt[i] + 1)) % phi;
  long long ans = 1ll;
  for (int i = 1; i < N; i++) {
    if (!cnt[i]) continue;
    long long now =
        binpow(i, (1ll * cnt[i] * (cnt[i] + 1)) / 2, (long long)(1e9 + 7));
    ans = (ans * binpow(now, (suff[i + 1] * pref[i - 1]) % phi,
                        (long long)(1e9 + 7))) %
          (long long)(1e9 + 7);
  }
  cout << ans << endl;
  return 0;
}
