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;
int i, j, k, n, m, ma;
int p[1000010], f[1000010 * 10], a[1000010], s[1000010 * 10];
long long g[1000010], an;
inline void prepare() {
  int i, j;
  for (i = 2; i <= ma; i++) {
    if (!f[i]) p[++*p] = i, f[i] = *p;
    for (j = 1; j <= f[i] && i * p[j] <= ma; j++) f[i * p[j]] = j;
  }
  for (i = 1; i <= n; i++) s[a[i]]++;
  for (i = ma; i; i--) s[i] += s[i + 1];
  for (i = 1; i <= ma; i++) {
    for (j = i; j > 1; j /= p[f[j]]) g[f[j]] += s[i];
  }
}
inline void work() {
  int i;
  long long l, r, mid, s, k;
  an = 1;
  for (i = 1; i <= *p; i++) {
    l = 1, r = (long long)1e14;
    for (; l <= r;) {
      mid = (l + r) / 2;
      for (s = 0, k = mid / p[i]; k; k /= p[i]) s += k;
      if (s < g[i])
        l = mid + 1;
      else
        r = mid - 1;
    }
    an = max(an, r + 1);
  }
  printf("%I64d\n", an);
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]), ma = max(ma, a[i]);
  prepare();
  work();
  return 0;
}