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 N = 2e5 + 5;
const long long Mod = 1e9 + 7;
long long n, m, a[N], p[N];
long long power(long long x, long long p, long long MOD) {
  long long s = 1;
  while (p) {
    if (p & 1) {
      s = (s * x) % MOD;
    }
    x = (x * x) % MOD;
    p >>= 1;
  }
  return s;
}
int main(int argc, char const *argv[]) {
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) {
    int x;
    scanf("%d", &x);
    ++p[x];
  }
  for (int i = 2; i < N; ++i) {
    if (p[i]) {
      a[++n] = i;
      p[n] = p[i];
    }
  }
  long long ans = 1, dx = 1, fx;
  for (int i = 1; i <= n; ++i) {
    fx = power(a[i], (p[i] * (p[i] + 1) / 2), Mod);
    ans = (power(ans, p[i] + 1, Mod) * power(fx, dx, Mod)) % Mod;
    dx = (dx * (p[i] + 1)) % (Mod - 1);
  }
  printf("%lld", ans);
  return 0;
}
