Problem: In a far away galaxy there is war again. The treacherous Republic made k precision strikes of power ai on the Empire possessions. To cope with the republican threat, the Supreme Council decided to deal a decisive blow to the enemy forces. 

To successfully complete the conflict, the confrontation balance after the blow should be a positive integer. The balance of confrontation is a number that looks like <image>, where p = n! (n is the power of the Imperial strike), <image>. After many years of war the Empire's resources are low. So to reduce the costs, n should be a minimum positive integer that is approved by the commanders.

Help the Empire, find the minimum positive integer n, where the described fraction is a positive integer.

Input

The first line contains integer k (1 ≤ k ≤ 106). The second line contains k integers a1, a2, ..., ak (1 ≤ ai ≤ 107).

Output

Print the minimum positive integer n, needed for the Empire to win.

Please, do not use the %lld to read or write 64-but integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.

Examples

Input

2
1000 1000


Output

2000

Input

1
2


Output

2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = int(1e7) + 5;
int pr[N];
vector<int> primes;
unsigned long long P[N];
bool PODHODIT(unsigned long long n) {
  for (int i = 0; i < primes.size(); i++) {
    unsigned long long sn = n;
    unsigned long long counter = 0;
    while (sn) {
      sn /= primes[i];
      counter += sn;
    }
    if (counter < P[primes[i]]) return false;
  }
  return true;
}
int main() {
  ios::sync_with_stdio(false);
  for (int i = 2; i <= N; i++) {
    if (!pr[i]) {
      primes.push_back(i);
      pr[i] = i;
    }
    for (int j = 0;
         j < primes.size() && i * primes[j] <= N && pr[i] >= primes[j]; j++)
      pr[i * primes[j]] = primes[j];
  }
  int n;
  cin >> n;
  unsigned long long rg = 0, lg = 0;
  for (int i = 0; i < n; i++) {
    int inp;
    cin >> inp;
    rg += inp;
    P[inp]++;
  }
  for (int i = N - 2; i >= 2; i--) P[i] += P[i + 1];
  for (int i = N - 1; i >= 2; i--) {
    if (pr[i] == i) continue;
    P[pr[i]] += P[i];
    P[i / pr[i]] += P[i];
    P[i] = 0;
  }
  while (lg < rg) {
    unsigned long long mid = (lg + rg) / 2;
    if (PODHODIT(mid))
      rg = mid;
    else
      lg = mid + 1;
  }
  cout << (rg ? rg : 1) << endl;
  return 0;
}