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;
long long d[11000000], a[11000000];
bool v[11000000];
long long ans, sum;
int n;
long long gao(long long x, long long y) {
  long long l = x, r = sum, mid, now, tmp;
  while (l <= r) {
    mid = (l + r) / 2;
    now = 0;
    tmp = mid;
    while (tmp > 0) {
      now += tmp / x;
      tmp /= x;
    }
    if (now >= y)
      r = mid - 1;
    else
      l = mid + 1;
  }
  return r + 1;
}
int main() {
  long long i, j, k;
  int x;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    sum += x;
    d[2]++;
    d[x + 1]--;
  }
  partial_sum(d, d + 11000000, d);
  for (i = 2; i < 11000000; i++)
    if (!v[i])
      for (j = i; j < 11000000; j *= i)
        for (k = j; k < 11000000; k += j) {
          v[k] = true;
          a[i] += d[k];
        }
  ans = 1;
  for (i = 2; i < 11000000; i++)
    if (a[i] > 0) ans = max(ans, gao(i, a[i]));
  printf("%I64d", ans);
}