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 INF = 1000000000;
const long long INFll = 1ll * INF * INF;
const int MOD = 1000000007;
const int MAXN = 200001;
int k[MAXN];
int bpow(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = (1ll * res * a) % MOD;
    b >>= 1;
    a = (1ll * a * a) % MOD;
  }
  return res;
}
int main() {
  int m, n = 1;
  scanf("%d", &m);
  for (int i = 0; i < m; ++i) {
    int p;
    scanf("%d", &p);
    n = (1ll * n * p) % MOD;
    ++k[p];
  }
  bool sqr = true;
  for (int i = 2; i < MAXN; ++i) sqr &= (k[i] % 2 == 0);
  int ans;
  if (!sqr) {
    bool f = false;
    ans = n;
    for (int i = 2; i < MAXN; ++i) {
      if (!k[i]) continue;
      if (!f && (k[i] % 2)) {
        ans = bpow(ans, (k[i] + 1) / 2);
        f = true;
      } else
        ans = bpow(ans, k[i] + 1);
    }
  } else {
    ans = 1;
    for (int i = 2; i < MAXN; ++i)
      if (k[i]) ans = (1ll * ans * bpow(i, k[i] / 2)) % MOD;
    for (int i = 2; i < MAXN; ++i)
      if (k[i]) ans = bpow(ans, k[i] + 1);
  }
  cout << ans << endl;
  return 0;
}
