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, syc;
long long ar[1000005], ans;
long long cnt[1000005 * 10];
bool p[1000005 * 10];
void read() {
  long long t, i;
  cin >> n;
  for (i = 1; i <= n; i++) scanf("%I64d", ar + i), cnt[ar[i]]++;
  return;
}
void init() {
  for (int i = 1e7; i >= 0; i--) cnt[i] += cnt[i + 1];
}
void solve() {
  long long bas, son, ort, t, sy, i, j, k;
  read();
  init();
  sort(ar + 1, ar + n + 1);
  ans = 1;
  for (i = 2; i <= ar[n]; i++)
    if (!p[i]) {
      bas = 1;
      son = 1LL << 62;
      syc = 0;
      for (k = i; k <= ar[n]; k *= i) {
        for (j = k; j <= ar[n]; j += k, sy++) {
          syc += cnt[j];
          p[j] = 1;
        }
      }
      while (bas < son) {
        ort = (bas + son) / 2;
        t = ort;
        sy = 0;
        while (t) sy += t / i, t /= i;
        if (sy >= syc)
          son = ort;
        else
          bas = ort + 1;
      }
      ans = ((ans) > (bas) ? (ans) : (bas));
    }
  cout << ans << endl;
  return;
}
int main() {
  solve();
  return 0;
}