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;
#pragma comment(linker, "/STACK:102400000,102400000")
long long pre[200020], suf[200020];
int a[200020];
int qpow(int x, int k) {
  int ret = 1;
  while (k) {
    if (k & 1) ret = 1LL * ret * x % 1000000007;
    x = 1LL * x * x % 1000000007;
    k >>= 1;
  }
  return ret;
}
int main() {
  int n;
  scanf("%d", &n);
  int mx = 0;
  for (int i = 0; i < n; ++i) {
    int v;
    scanf("%d", &v);
    a[v]++;
    mx = max(mx, v);
  }
  pre[0] = suf[mx + 1] = 1;
  for (int i = 1; i <= mx; ++i)
    pre[i] = pre[i - 1] * (a[i] + 1) % (1000000007 - 1);
  for (int i = mx; i > 0; --i)
    suf[i] = suf[i + 1] * (a[i] + 1) % (1000000007 - 1);
  long long ans = 1;
  for (int i = 2; i <= mx; ++i) {
    if (a[i]) {
      long long k = pre[i - 1] * suf[i + 1] % (1000000007 - 1);
      k = 1LL * (a[i] + 1) * a[i] / 2 % (1000000007 - 1) * k % (1000000007 - 1);
      ans = ans * qpow(i, k) % 1000000007;
    }
  }
  cout << ans << endl;
  return 0;
}
