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>
#pragma GCC optimize("O3")
using namespace std;
const int N = 1e7 + 5;
int mx;
int arr[N];
int cnt[N];
bool notp[N];
map<int, long long> primes;
void sieve() {
  for (int i = 2; i <= mx; ++i) {
    if (notp[i]) continue;
    long long &res = primes[i];
    long long v = (i);
    for (long long j = i; j <= mx; j += i) {
      notp[j] = 1;
      if (j == v) {
        for (long long k = v; k <= mx; k += v) {
          res += cnt[k];
        }
        v *= i;
      }
    }
  }
}
bool valid(long long v) {
  for (auto &z : primes) {
    long long x = z.first;
    long long cnt = 0;
    long long t = v;
    while (t) {
      cnt += t / x;
      t /= x;
    }
    if (cnt < z.second) return 0;
  }
  return 1;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", arr + i);
    mx = max(mx, arr[i]);
    cnt[arr[i]]++;
  }
  for (int i = mx; i; --i) {
    cnt[i] += cnt[i + 1];
  }
  sieve();
  long long s = mx, e = 1e14 + 5;
  while (s != e) {
    long long mid = s + (e - s) / 2;
    if (valid(mid))
      e = mid;
    else
      s = mid + 1;
  }
  printf("%lld", s);
}