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;
int mod = 1000000007;
int a[200005];
map<int, int> cnt;
long long getPow(long long x, long long p, long long mod) {
  if (p == 0) return 1;
  long long ans = getPow(x, p / 2, mod);
  ans = (ans * ans) % mod;
  if (p % 2) ans = (ans * x) % mod;
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    cnt[a[i]]++;
  }
  long long product = 1, num = 1;
  for (map<int, int>::iterator it = cnt.begin(); it != cnt.end(); it++) {
    long long val = it->first;
    long long rep = it->second;
    long long inc = 0, mul = 1;
    for (int i = 1; i <= rep; i++) {
      inc += num;
      inc %= (mod - 1);
      mul = (((mul * product) % mod) * getPow(val, i * num, mod)) % mod;
    }
    num = (num + inc) % (mod - 1);
    product = (product * mul) % mod;
  }
  cout << product << endl;
  return 0;
}
