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 + 10;
const int MAXM = 1e5 + 10;
const int INF = 1e9;
const int MOD = 1e9 + 7;
int min(int a, int b) { return a < b ? a : b; }
int max(int a, int b) { return a > b ? a : b; }
long long cnt[MAXN];
int lp[MAXN], pm[MAXN], Nump;
void init() {
  for (int i = 2; i < MAXN; i++) {
    if (!lp[i]) {
      lp[i] = i;
      pm[Nump++] = i;
      for (int j = i * 2; j < MAXN; j += i)
        if (!lp[j]) lp[j] = i;
    }
  }
  for (int i = MAXN - 2; i >= 2; i--) cnt[i] += cnt[i + 1];
  for (int i = MAXN - 2; i >= 2; i--) {
    if (lp[i] != i) {
      cnt[lp[i]] += cnt[i];
      cnt[i / lp[i]] += cnt[i];
    }
  }
}
bool Test(long long x) {
  for (int i = 0; i < Nump; i++)
    if (cnt[pm[i]]) {
      int p = pm[i];
      long long sum = 0, xx = x;
      while (xx /= p) sum += xx;
      if (sum < cnt[p]) return false;
    }
  return true;
}
void solve() {
  init();
  long long lf = 0, rt = 1e13;
  while (lf + 1 < rt) {
    long long m = (lf + rt) >> 1;
    if (Test(m))
      rt = m;
    else
      lf = m;
  }
  cout << rt << endl;
}
int main() {
  int k;
  scanf("%d", &k);
  for (int i = 0; i < k; i++) {
    int x;
    scanf("%d", &x);
    cnt[x]++;
  }
  solve();
  return 0;
}