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>
const long long p = 1e9 + 7;
const long long p2 = (p - 1) * 2;
int n;
int a[200010];
int vis[200010];
int cnt[200010];
long long ans, tot;
long long ksm(long long di, long long zhi, long long p) {
  long long res = 1;
  while (zhi > 0) {
    if (zhi & 1) res = (res * di) % p;
    di = (di * di) % p;
    zhi >>= 1;
  }
  return res;
}
int main() {
  while (scanf("%d", &n) != EOF) {
    ans = 1;
    memset(vis, 0, sizeof(vis));
    memset(cnt, 0, sizeof(cnt));
    for (int i = 0; i < n; i++) {
      scanf("%d", &a[i]);
      cnt[a[i]]++;
    }
    tot = 1;
    for (int i = 0; i < n; i++) {
      if (vis[a[i]]) continue;
      vis[a[i]] = 1;
      tot = (tot * (cnt[a[i]] + 1)) % p2;
    }
    memset(vis, 0, sizeof(vis));
    for (int i = 0; i < n; i++) {
      if (vis[a[i]]) continue;
      vis[a[i]] = 1;
      ans = (ans * ksm(a[i], (tot * cnt[a[i]] / 2) % p2, p)) % p;
    }
    printf("%lld\n", ans);
  }
  return 0;
}
