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 x;
long long a[200500], n, d[200500], now;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long Pow(long long x, long long y) {
  long long ans = 1;
  while (y) {
    if (y & 1) ans = ans * x % 1000000007;
    x = x * x % 1000000007;
    y >>= 1;
  }
  return ans;
}
int main() {
  n = read();
  for (long long i = 1; i <= n; i++) {
    x = read();
    d[x]++;
  }
  long long ans = 1, D = 1;
  for (long long i = 2; i <= 200000; i++)
    if (d[i]) {
      long long fb = Pow(i, d[i] * (d[i] + 1) / 2) % 1000000007;
      ans = Pow(ans, d[i] + 1) * Pow(fb, D) % 1000000007;
      D = D * (d[i] + 1) % (1000000007 - 1);
    }
  printf("%lld\n", ans);
  return 0;
}
