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 mo = 1000000007;
long long a, ans, zheng[210000], fan[210000], num[210000], m, i;
long long search(long long a, long long b) {
  long long c;
  if (b == 1) return a;
  c = search(a * a % mo, b / 2);
  if (b % 2 == 1) c = c * a % mo;
  return c;
}
int main() {
  scanf("%lld", &m);
  for (i = 1; i <= m; i++) scanf("%lld", &a), num[a]++;
  zheng[1] = 1;
  fan[200001] = 1;
  fan[1] = 1;
  zheng[200001] = 1;
  for (i = 2; i <= 200000; i++)
    zheng[i] = zheng[i - 1] * (num[i] + 1) % (mo - 1);
  for (i = 200000; i >= 2; i--) fan[i] = fan[i + 1] * (num[i] + 1) % (mo - 1);
  ans = 1;
  for (i = 2; i <= 200000; i++)
    if (num[i] != 0)
      ans = ans *
            search(i, ((fan[i + 1] * zheng[i - 1] % (mo - 1)) *
                       (((num[i] + 1) * num[i] / 2) % (mo - 1))) %
                          (mo - 1)) %
            mo;
  printf("%lld\n", ans);
}
