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 nums[200000];
inline int mod(int x, int y, int m) {
  long long z = (x * 1LL * y);
  if (z >= m) return z % m;
  return (int)z;
}
int modexpo(int x, int y) {
  if (!y) return 1;
  if (y & 1) return mod(x, modexpo(mod(x, x, 1e9 + 7), y / 2), 1e9 + 7);
  return modexpo(mod(x, x, 1e9 + 7), y / 2);
}
int main() {
  int i, j, k, l, m, n, x;
  scanf("%d", &m);
  for (i = 0; i < m; i++) scanf("%d", &x), nums[x]++;
  int ans = 1;
  int c = 0;
  bool check = false;
  for (i = 0; i < 200000; i++)
    if (nums[i] & 1) c++;
  if (c) {
    for (i = 0; i < 200000; i++) {
      if (check)
        ans = mod(ans, nums[i] + 1, 1e9 + 6);
      else if (nums[i] & 1) {
        ans = mod(ans, (nums[i] + 1) / 2, 1e9 + 6);
        check = true;
      } else
        ans = mod(ans, (nums[i] + 1), 1e9 + 6);
    }
    int n = 1;
    for (i = 2; i < 200000; i++) n = mod(n, modexpo(i, nums[i]), 1e9 + 7);
    printf("%d ", modexpo(n, ans));
  } else {
    for (i = 0; i < 200000; i++)
      ans = mod(ans, (nums[i] + 1), 1e9 + 6), nums[i] /= 2;
    int n = 1;
    for (i = 2; i < 200000; i++) n = mod(n, modexpo(i, nums[i]), 1e9 + 7);
    printf("%d ", modexpo(n, ans));
  }
  return 0;
}
