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>
const int N = (int)1e7 + 100;
int lp[N], simples[N], simsz;
long long cnt[N];
bool ok(long long mid) {
  bool can = true;
  for (int i = 0; i < simsz && can; i++) {
    long long tmp = mid, f = 0;
    while (tmp) {
      tmp /= simples[i];
      f += tmp;
    }
    if (f < cnt[i]) can = false;
  }
  return can;
}
int main() {
  int n, max = 0;
  scanf("%d", &n);
  long long rg = 0, lf = 1;
  for (int i = 0; i < n; ++i) {
    int a;
    scanf("%d", &a);
    if (a > max) max = a;
    rg += a;
    ++cnt[a];
  }
  for (int i = 2; i <= max; ++i) {
    if (lp[i] == 0) {
      lp[i] = i;
      simples[simsz++] = i;
    }
    for (int j = 0;
         j < simsz && simples[j] <= lp[i] && (long long)i * simples[j] <= max;
         ++j) {
      lp[i * simples[j]] = simples[j];
    }
  }
  for (int i = max; i >= 2; i--) cnt[i] += cnt[i + 1];
  for (int i = max; i >= 2; i--) {
    if (lp[i] != i) cnt[lp[i]] += cnt[i];
    cnt[i / lp[i]] += cnt[i];
  }
  for (int i = 0; i < simsz; i++) {
    cnt[i] = cnt[simples[i]];
  }
  while (rg != lf) {
    long long mid = (rg + lf) >> 1;
    if (ok(mid)) {
      rg = mid;
    } else
      lf = mid + 1;
  }
  printf("%I64d", lf);
}