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;
long long mpre[100100], mpos[100100];
int n;
map<long long, long long> m;
map<long long, long long>::iterator it;
long long fpow(long long a, long long b) {
  long long acm = 1;
  if (!b) return acm;
  acm = fpow(a, b / 2);
  acm = (acm * acm) % 1000000007;
  if (b % 2) acm = (acm * a) % 1000000007;
  return acm % 1000000007;
}
long long fsum(long long num) {
  return ((((num + 1) * num) / 2) % (1000000007 - 1));
}
int main() {
  cin >> n;
  long long a;
  for (int i = 0; i < n; i++) {
    scanf("%I64d", &a);
    m[a]++;
  }
  for (int i = 0; i < 100100; i++) {
    mpre[i] = 1;
    mpos[i] = 1;
  }
  int ind = 1;
  for (it = m.begin(); it != m.end(); it++, ind++) {
    mpre[ind] = (mpre[ind - 1] * (it->second + 1)) % (1000000007 - 1);
  }
  for (it--, ind--;; it--, ind--) {
    mpos[ind] = (mpos[ind + 1] * (it->second + 1)) % (1000000007 - 1);
    if (it == m.begin()) break;
  }
  long long acm, res = 1;
  ind = 1;
  for (it = m.begin(); it != m.end(); it++, ind++) {
    acm = (fsum(it->second) *
           ((mpre[ind - 1] * mpos[ind + 1]) % (1000000007 - 1))) %
          (1000000007 - 1);
    res = (res * fpow(it->first, acm)) % 1000000007;
  }
  cout << res % 1000000007 << endl;
  return 0;
}
