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 MOD = 1e9 + 7;
long long power(long long b, long long e) {
  long long ret = 1;
  while (e) {
    if (e % 2 == 1) {
      ret = (ret * b) % MOD;
      --e;
    }
    e /= 2;
    b = (b * b) % MOD;
  }
  return ret;
}
int main() {
  int n, tmp = -1;
  scanf("%d", &n);
  map<int, int> freq;
  long long base = 1;
  for (int i = 0; i < n; ++i) {
    int a;
    scanf("%d", &a);
    if (tmp == -1) tmp = a;
    ++freq[a];
    base = (base * a) % MOD;
  }
  bool even = false;
  long long rootBase = 1;
  for (auto &p : freq) {
    ++p.second;
    rootBase = (rootBase * power(p.first, p.second / 2)) % MOD;
    if (p.second % 2 == 0) even = true;
  }
  bool divided = false;
  for (auto p : freq) {
    if (even) {
      if (p.second % 2 == 0 && !divided) {
        divided = true;
        p.second /= 2;
      }
    } else if (!divided) {
      divided = true;
      p.second /= 2;
      base = power(base, p.second);
      base = (base * rootBase) % MOD;
      continue;
    }
    base = power(base, p.second);
  }
  printf("%lld\n", base);
  return 0;
}
