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 = 1e7 + 5;
long long power[N];
bool mark[N];
int cnt[N];
inline long long count(long long x, int k) {
  long long ans = 0;
  while (x >= k) {
    ans += x / k;
    x /= k;
  }
  return ans;
}
inline void readInput() {
  int k;
  cin >> k;
  for (int i = 0; i < k; i++) {
    int x;
    cin >> x;
    cnt[x]++;
  }
}
inline void solve() {
  for (int i = N - 1; i; i--) cnt[i - 1] += cnt[i];
  for (int i = 2; i < N; i++)
    if (!mark[i])
      for (int j = i; j < N; j += i) {
        mark[j] = true;
        if (cnt[j]) {
          int p = 0;
          for (int k = j; !(k % i); k /= i) p++;
          power[i] += cnt[j] * p;
        }
      }
  long long ans = 1;
  for (int i = 1e7; i > 1; i--) {
    if (!power[i] || count(ans, i) >= power[i]) continue;
    long long L = ans, R = 1e15;
    while (L + 1 < R) {
      long long mid = L + R >> 1;
      (count(mid, i) < power[i] ? L : R) = mid;
    }
    ans = R;
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  readInput(), solve();
  return 0;
}