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>
long long pow(int a, int b) {
  long long x = 1, y = a;
  while (b) {
    if (b & 1) {
      x = x * y % 1000000007;
    }
    y = y * y % 1000000007;
    b /= 2;
  }
  return x;
}
int freq[200005];
int main() {
  int n, k;
  scanf("%d", &n);
  long long p = 1;
  for (int i = 0; i < n; i++) {
    scanf("%d", &k);
    freq[k]++;
  }
  int alleven = true;
  for (int i = 1; i <= 200000; i++) {
    alleven &= (freq[i] % 2 == 0);
  }
  long long x = 1;
  for (int i = 1; i <= 200000; i++) {
    int k = freq[i];
    if (alleven) {
      k /= 2;
    }
    for (int j = 0; j < k; j++) {
      x *= i;
      x %= 1000000007;
    }
  }
  bool divided = alleven;
  for (int i = 1; i <= 200000; i++) {
    int p;
    if (!divided && freq[i] % 2 == 1) {
      p = (freq[i] + 1) / 2;
      divided = true;
    } else {
      p = freq[i] + 1;
    }
    x = pow(x, p);
  }
  printf("%I64d\n", x);
  return 0;
}
