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 p[10000100], flag[10000100], nn, a[1000100], ct[10000100];
long long cnt[10000100];
void pre() {
  for (int i = 2; i < 10000100; i++) {
    if (flag[i]) continue;
    p[++nn] = i;
    for (int j = i; j < 10000100; j += i) flag[j] = i;
  }
}
int main() {
  pre();
  int k;
  cin >> k;
  int mx = 0;
  for (int i = 1; i <= k; i++) {
    scanf("%d", &a[i]);
    mx = max(mx, a[i]);
    ct[a[i]]++;
  }
  for (int i = 1; i <= mx; i++) ct[i] += ct[i - 1];
  int id = upper_bound(p + 1, p + nn + 1, mx) - p - 1;
  for (int i = 1; i <= id; i++) {
    for (long long t = p[i]; t <= mx; t *= p[i]) {
      for (int k = 1; k <= mx / t; k++) {
        cnt[i] +=
            1LL * k * (ct[min((long long)mx, t * k + t - 1)] - ct[t * k - 1]);
      }
    }
  }
  long long st = 0, ed = 0;
  for (int i = 1; i <= k; i++) ed += a[i];
  while (st + 1 < ed) {
    long long md = (st + ed) >> 1;
    int flag = 0;
    for (int i = 1; i <= id; i++) {
      long long t = 0, q = p[i];
      long long tmp = md;
      while (tmp >= q) {
        t += tmp / q;
        tmp /= q;
      }
      if (t < cnt[i]) {
        flag = 1;
        break;
      }
    }
    if (flag)
      st = md;
    else
      ed = md;
  }
  cout << ed << endl;
  return 0;
}