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;
long long n, ans = 1, sum;
long long ar[1000002];
long long cnt[10000002];
bool pr[10000002];
int main() {
  scanf("%I64d", &n);
  for (long long i = 1; i <= n; i++) {
    scanf("%I64d", &ar[i]);
    cnt[ar[i]]++;
    sum += ar[i];
  }
  sort(ar + 1, ar + n + 1);
  for (long long i = 10000002 - 2; i >= 0; i--) cnt[i] += cnt[i + 1];
  for (long long i = 2; i <= ar[n]; i++)
    if (!pr[i]) {
      long long tot = 0;
      for (long long j = i; j <= ar[n]; j *= i)
        for (long long k = j; k <= ar[n]; k += j) {
          tot += cnt[k];
          pr[k] = true;
        }
      long long left = ans, right = sum, last;
      while (left <= right) {
        long long mid = (left + right) >> 1;
        long long tot2 = 0;
        long long tmp = mid;
        while (tmp) {
          tot2 += tmp / i;
          tmp /= i;
        }
        if (tot2 >= tot) {
          last = mid;
          right = mid - 1;
        } else
          left = mid + 1;
      }
      ans = last;
    }
  printf("%I64d\n", ans);
  return 0;
}