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 + 6;
int low[N], isp[N];
vector<int> pri;
void getprime() {
  fill_n(isp + 2, N - 2, 1);
  for (int i = (2); i < (N); ++i) {
    if (isp[i]) pri.push_back(i);
    for (auto p : pri) {
      if (p * i >= N) break;
      low[p * i] = p;
      isp[p * i] = 0;
      if (i % p == 0) break;
    }
  }
}
int n, a[N];
long long cnt[N];
bool ck(long long mid) {
  for (auto p : pri) {
    long long ned = cnt[p];
    for (long long tmp = mid / p; tmp; tmp /= p) ned -= tmp;
    if (ned > 0) return 0;
  }
  return 1;
}
int main() {
  getprime();
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) {
    scanf("%d", a + i);
    cnt[a[i]]++;
  }
  for (int i = (N - 1) - 1; i >= (2); --i) cnt[i] += cnt[i + 1];
  for (int i = (N - 1) - 1; i >= (2); --i)
    if (!isp[i]) {
      cnt[low[i]] += cnt[i];
      cnt[i / low[i]] += cnt[i];
    }
  long long l = 1, r = 0, ans;
  for (int i = (0); i < (n); ++i) r += a[i];
  for (long long mid = l + r >> 1; l <= r; mid = l + r >> 1)
    ck(mid) ? r = (ans = mid) - 1 : l = mid + 1;
  printf("%lld\n", ans);
  return 0;
}