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 INF = 1001001001;
const unsigned long long INFLL = 1001001001001001001LL;
template <typename T>
void pv(T a, T b) {
  for (T i = a; i != b; ++i) cout << *i << " ";
  cout << endl;
}
template <typename T>
void chmin(T& a, T b) {
  if (a > b) a = b;
}
template <typename T>
void chmax(T& a, T b) {
  if (a < b) a = b;
}
const unsigned long long MAX = 10000050;
int occ[MAX] = {}, isp[MAX];
int main() {
  int k;
  scanf("%d", &k);
  for (int i = 0; i < k; ++i) {
    int a;
    scanf("%d", &a);
    ++occ[a];
  }
  for (int i = 1; i < MAX; ++i) {
    occ[i] += occ[i - 1];
  }
  for (int i = 2; i < MAX; ++i) {
    isp[i] = 1;
  }
  unsigned long long n = 1;
  for (int i = 2; i < MAX; ++i) {
    if (isp[i]) {
      for (int j = i + i; j < MAX; j += i) {
        isp[j] = 0;
      }
      unsigned long long cnt = 0;
      for (unsigned long long p = i; p < MAX; p *= i) {
        for (unsigned long long q = p; q - 1 < MAX; q += p) {
          cnt += k - occ[q - 1];
        }
      }
      unsigned long long lo = 0, hi = 10000000000000LL;
      while (hi - lo > 1) {
        unsigned long long mid = (hi + lo) / 2;
        unsigned long long cc = 0;
        for (unsigned long long p = i; p <= mid; p *= i) {
          cc += mid / p;
        }
        if (cc >= cnt) {
          hi = mid;
        } else {
          lo = mid;
        }
      }
      chmax(n, hi);
    }
  }
  cout << n << endl;
  return 0;
}