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> prime;
vector<int> f;
void Init_prime(int N) {
  prime.clear();
  f = vector<int>(N + 1, 0);
  for (int i = 2; i <= N; i++) {
    if (f[i]) {
      for (int j = 0; j < prime.size() && prime[j] * i <= N; j++) {
        f[prime[j] * i] = prime[j];
        if (i % prime[j] == 0) break;
      }
    } else {
      f[i] = i;
      prime.push_back(i);
      for (int j = 0; j < prime.size() && prime[j] * i <= N; j++)
        f[prime[j] * i] = prime[j];
    }
  }
}
int a[1000000];
long long b[10000001];
int main() {
  int n;
  scanf("%d", &n);
  long long mx = 0, mn = 1;
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    mx += a[i];
    b[a[i]]++;
  }
  sort(a, a + n);
  Init_prime(a[n - 1]);
  for (int i = a[n - 1] - 1; i >= 2; i--) b[i] += b[i + 1];
  vector<long long> cnt = vector<long long>(a[n - 1] + 1, 0);
  for (int i = a[n - 1]; i >= 2; i--) {
    cnt[f[i]] += b[i];
    if (f[i] != i) b[i / f[i]] += b[i];
  }
  long long mid, tmp;
  while (mn <= mx) {
    mid = mn + mx >> 1;
    bool ok = 1;
    for (int i = 0; i < prime.size(); i++) {
      long long _cnt = 0;
      tmp = mid;
      while (tmp) {
        tmp /= prime[i];
        _cnt += tmp;
      }
      if (_cnt < cnt[prime[i]]) {
        ok = 0;
        break;
      }
    }
    if (ok)
      mx = mid - 1;
    else
      mn = mid + 1;
  }
  printf("%I64d\n", mx + 1);
}