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;
vector<int> pr;
long long cnt[11111111];
int lp[11111111];
int N = 10000000;
bool check(long long a) {
  for (int i = 0; i < pr.size(); ++i) {
    if (!cnt[pr[i]]) break;
    long long kol = 0;
    long long T = a;
    do {
      kol += T / pr[i];
      T /= pr[i];
    } while (T >= pr[i]);
    if (kol < cnt[pr[i]]) return 0;
  }
  return 1;
}
int g[11111111];
int main() {
  for (int i = 2; i <= N; ++i) {
    if (!lp[i]) {
      lp[i] = i;
      pr.push_back(i);
    }
    long long c;
    for (int j = 0;
         j < int(pr.size()) && pr[j] <= lp[i] && (c = i * pr[j]) <= N; ++j)
      lp[c] = pr[j];
  }
  int i, j, k, n;
  cin >> n;
  for (i = 0; i < n; ++i) {
    cin >> k;
    g[k + 1]--;
  }
  long long flag = n;
  for (i = 1; i <= N; ++i) {
    flag += g[i];
    cnt[i] = flag;
  }
  for (i = N; i > 1; --i) {
    if (lp[i] != i) cnt[lp[i]] += cnt[i];
    cnt[i / lp[i]] += cnt[i];
  }
  long long mi, lo = 1, hi = 11000000000000LL;
  for (; lo < hi;) {
    mi = (lo + hi) / 2ll;
    if (check(mi))
      hi = mi;
    else
      lo = mi + 1;
  }
  cout << lo;
  return 0;
}