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 + 20;
bool vis[N];
int tot;
int prime[N];
long long cnt[N], a[N];
void shai(int x) {
  for (int i = 2; i <= x; i++) {
    if (!vis[i]) prime[++tot] = i;
    for (int j = 1; j <= tot && 1LL * i * prime[j] <= x; j++) {
      vis[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
bool check(long long mid) {
  long long now;
  for (int i = 1; i <= tot; i++) {
    now = 0;
    for (long long j = prime[i]; j <= mid; j *= prime[i]) now += mid / j;
    if (now < cnt[i]) return 0;
  }
  return 1;
}
int main() {
  int n;
  scanf("%d", &n);
  long long l = 1, r = 0, mid;
  int x, maxx = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    a[x]++;
    maxx = max(maxx, x);
    r += x;
  }
  for (int i = maxx; i >= 1; i--) a[i] += a[i + 1];
  shai(maxx);
  for (int i = 1; i <= tot; i++)
    for (long long j = prime[i]; j <= maxx; j *= prime[i])
      for (long long k = j; k <= maxx; k += j) cnt[i] += a[k];
  while (l < r) {
    mid = (l + r) / 2;
    if (check(mid))
      r = mid;
    else
      l = mid + 1;
  }
  printf("%lld\n", r);
  return 0;
}