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 maxn = 1e7;
int n, tot, vis[maxn + 1];
int a[1000002], p[maxn + 1];
long long sum, cnt[maxn + 1];
long long res[maxn + 1];
bool judge(long long mid) {
  memset(res, 0, sizeof(res));
  for (int i = 1; i <= tot; i++) {
    long long t = p[i];
    while (mid / t) {
      res[p[i]] += mid / t;
      t *= p[i];
    }
  }
  for (int i = 1; i <= tot; i++) {
    if (cnt[p[i]] > res[p[i]]) return false;
  }
  return true;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), cnt[a[i]]++, sum += a[i];
  for (int i = maxn; i >= 2; i--) cnt[i - 1] += cnt[i];
  for (int i = 2; i <= maxn; i++) {
    if (!vis[i]) p[++tot] = i, vis[i] = i;
    for (int j = 1; j <= tot && p[j] * i <= maxn; j++) {
      vis[i * p[j]] = p[j];
      if (i % p[j] == 0) break;
    }
  }
  for (int i = maxn; i >= 2; i--)
    if (vis[i] != i) {
      cnt[i / vis[i]] += cnt[i];
      cnt[vis[i]] += cnt[i];
    }
  long long l = 0, r = sum;
  long long ans = 0;
  while (l + 1 < r) {
    long long mid = (l + r) >> 1;
    if (judge(mid))
      r = mid;
    else
      l = mid;
  }
  printf("%I64d\n", r);
  return 0;
}