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 = 200200;
const int M = 1e9 + 7;
int n;
int a[N];
int num[N], c[N], b[N];
int bp(int a, int b) {
  int r = 1;
  while (b > 0) {
    if (b % 2 == 1) r = 1LL * r * a % M;
    a = 1LL * a * a % M;
    b /= 2;
  }
  return r;
}
int main() {
  scanf("%d", &n);
  while (n--) {
    int x;
    scanf("%d", &x);
    ++a[x];
  }
  int nums = 0;
  for (int i = 0; i < N; ++i)
    if (a[i] > 0) {
      num[i] = nums;
      c[nums] = i;
      ++nums;
    }
  for (int i = 0; i < N; ++i) b[i] = i;
  int r = 1;
  int k = 1;
  for (int i = 0; i < N; ++i) {
    int p = 1, pr = r;
    for (int j = 0; j < a[i]; ++j) {
      p = 1LL * p * i % M;
      r = 1LL * r * pr % M * bp(p, k) % M;
    }
    k = 1LL * k * (a[i] + 1) % (M - 1);
  }
  printf("%d\n", r);
  return 0;
}
