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 long long maxn = 200000;
long long l[200010];
long long r[200010];
long long num[200010];
long long ans = 1;
int n, x;
long long quick_pow(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % 1000000007;
    b >>= 1;
    a = a * a % 1000000007;
  }
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    num[x]++;
  }
  l[0] = r[maxn + 1] = 1;
  for (int i = 1; i <= maxn; i++) {
    l[i] = l[i - 1] * (num[i] + 1ll) % (1000000007 - 1);
  }
  for (int i = maxn; i; i--) {
    r[i] = r[i + 1] * (num[i] + 1ll) % (1000000007 - 1);
  }
  for (int i = 1; i <= maxn; i++) {
    long long tem = l[i - 1] * r[i + 1] % (1000000007 - 1);
    tem = (tem) * (((1ll + num[i]) * num[i] / 2) % (1000000007 - 1)) %
          (1000000007 - 1);
    ans = (ans * quick_pow(i, tem)) % 1000000007;
  }
  cout << ans;
}
