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;
const int N = 11111111;
const int M = 666666;
int n, m, i, j, x, Max, p[M], mi[N];
long long q[N], s[N], tmp, l, r, mid, ans;
bool check(long long x) {
  for (int i = 0; i < m; i++) {
    long long t = 0;
    for (long long o = p[i]; o <= x; o *= p[i]) t += x / o;
    if (t < q[i]) return 0;
  }
  return 1;
}
int main() {
  scanf("%d", &n);
  for (; n--;) {
    scanf("%d", &x);
    s[x]++;
    Max = max(Max, x);
  }
  for (i = 2; i <= Max; i++) {
    if (!mi[i]) p[m++] = i;
    for (j = 0; j < m && i * p[j] <= Max; j++) {
      mi[i * p[j]] = p[j];
      if (i % p[j] == 0) break;
    }
  }
  for (i = Max; i >= 2; i--) {
    tmp += s[i];
    if (mi[i]) {
      s[mi[i]] += tmp;
      s[i / mi[i]] += tmp;
      s[mi[i] - 1] -= tmp;
      s[i / mi[i] - 1] -= tmp;
    } else
      q[lower_bound(p, p + m, i) - p] = tmp;
  }
  l = 1;
  r = ans = (long long)1 << (long long)60;
  while (l <= r) {
    mid = (l + r) / 2;
    if (check(mid)) {
      ans = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  printf("%I64d\n", ans);
}