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;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
const long long mod = 1000000007;
inline long long fpow(long long n, long long p) {
  n %= mod;
  long long ans = 1, base = n;
  while (p) {
    if (p & 1) ans = ans * base % mod;
    base = base * base % mod;
    p >>= 1;
  }
  return ans;
}
long long n, vis[200010], cj = 1, ans = 1;
vector<pair<long long, long long> > mp;
bool flag = 1;
signed main() {
  n = read();
  for (long long i = 1; i <= n; ++i) vis[read()]++;
  for (long long i = 1; i <= 200000; ++i) cj = cj * (vis[i] + 1) % (mod - 1);
  for (long long i = 1; i <= 200000; ++i)
    if (vis[i]) mp.push_back(make_pair(vis[i], i));
  for (long long i = 0; i < mp.size(); ++i) {
    if (mp[i].first & 1) {
      flag = 0;
      break;
    }
  }
  if (flag) {
    for (long long i = 0; i < mp.size(); ++i) mp[i].first >>= 1;
    for (long long i = 0; i < mp.size(); ++i)
      ans = ans * fpow(mp[i].second, cj * mp[i].first % (mod - 1)) % mod;
  } else {
    cj = 1;
    bool FI = 1;
    for (long long i = 0; i < mp.size(); ++i) {
      if (mp[i].first & 1 && FI) {
        FI = 0;
        cj = cj * (mp[i].first + 1) / 2 % (mod - 1);
      } else
        cj = cj * (mp[i].first + 1) % (mod - 1);
    }
    for (long long i = 0; i < mp.size(); ++i)
      ans = ans * fpow(mp[i].second, cj * mp[i].first % (mod - 1)) % mod;
  }
  printf("%lld", ans);
  return 0;
}
