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 MAX = 2e5 + 6;
const long long MOD = 1e9 + 7;
long long cnt[MAX];
long long pref[MAX], suf[MAX];
long long bigmod(long long x, long long p) {
  long long ret = 1LL;
  while (p) {
    if (p & 1) ret = (ret * x) % MOD;
    x = (x * x) % MOD;
    p >>= 1;
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  set<long long> st;
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    cnt[x]++;
    st.insert(x);
  }
  vector<long long> v(st.begin(), st.end());
  int k = v.size();
  pref[0] = suf[k + 1] = 1LL;
  for (int i = 1; i <= k; i++) {
    pref[i] = (pref[i - 1] * (cnt[v[i - 1]] + 1)) % (MOD - 1);
  }
  for (int i = k; i >= 1; i--) {
    suf[i] = (suf[i + 1] * (cnt[v[i - 1]] + 1)) % (MOD - 1);
  }
  long long ans = 1LL;
  for (int i = 1; i <= k; i++) {
    long long cur = 1LL;
    long long gun = (pref[i - 1] * suf[i + 1]) % (MOD - 1);
    for (int j = 0; j < cnt[v[i - 1]]; j++) {
      cur = (cur * v[i - 1]) % MOD;
      ans = (ans * bigmod(cur, gun)) % MOD;
    }
  }
  cout << ans;
}
