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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  vector<int> fact(1e7 + 1);
  for (int i = 2; i <= 3162; ++i)
    if (!fact[i])
      for (int j = i * i; j <= 1e7; j += i) fact[j] = i;
  int n, a;
  cin >> n;
  vector<long long> exp(1e7 + 2);
  while (n--) {
    cin >> a;
    ++exp[a];
  }
  for (int i = 1e7; i >= 2; --i) exp[i] += exp[i + 1];
  for (int i = 1e7; i >= 2; --i) {
    if (fact[i]) {
      exp[fact[i]] += exp[i];
      exp[i / fact[i]] += exp[i];
      exp[i] = 0;
    }
  }
  long long ans = 1;
  for (int i = 2; i <= 1e7; ++i) {
    if (ans >= exp[i] * i) continue;
    long long a = max(ans, exp[i] * (i - 1)) / i * i;
    long long b = 0;
    for (long long tmp = a / i; tmp > 0; tmp /= i) b += tmp;
    while (b < exp[i]) {
      a += i;
      ++b;
      for (long long tmp = a / i; tmp % i == 0; tmp /= i) ++b;
    }
    ans = max(ans, a);
  }
  cout << ans;
  return 0;
}