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;
void run();
int main() {
  ios::sync_with_stdio(0);
  run();
}
int n;
int a[1100000];
int pax[11000000];
long long need[11000000];
void run() {
  cin >> n;
  for (int i = n; i--;) cin >> a[i];
  sort(a, a + n);
  for (int i = 2; i <= a[n - 1]; i++) {
    if (not pax[i]) {
      pax[i] = i;
      if (a[n - 1] / i >= i) {
        for (int j = i; j <= a[n - 1]; j += i) {
          pax[j] = i;
        }
      }
    }
  }
  for (int i = a[n - 1], j = 0; i >= 2; i--) {
    while (j < n and a[n - 1 - j] == i) ++j;
    for (long long k = i; k != 1; k /= pax[k]) need[pax[k]] += j;
  }
  long long res = max(1ll, need[2]);
  for (int i = 2; i <= a[n - 1]; i++) {
    if (pax[i] == i) {
      long long tot = 0;
      for (long long q = i;; q *= i) {
        long long add = res / q;
        tot += add;
        if (add < i) break;
      }
      if (tot >= need[i]) continue;
      long long lef = res - 1, rad = 1ll << 42ll;
      while (rad >>= 1ll) {
        long long mid = lef + rad, tot = 0;
        if (mid > lef) {
          for (long long q = i;; q *= i) {
            long long add = mid / q;
            tot += add;
            if (add < i) break;
          }
          if (tot < need[i]) lef = mid;
        }
      }
      res = lef + 1;
    }
  }
  cout << res << endl;
}