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 = 10000010;
long long l, r;
int n, a[1000010], maxr, vis[N];
int prime[N], cnt, mindiv[N];
bool ok[N];
long long tot[N];
inline void init() {
  for (int i = 2; i <= maxr; i++) {
    if (!ok[i]) {
      prime[++cnt] = i;
      mindiv[i] = cnt;
    }
    for (int j = 1; j <= cnt && i * prime[j] <= maxr; j++) {
      ok[i * prime[j]] = 1;
      mindiv[i * prime[j]] = j;
      if (i % prime[j] == 0) break;
    }
  }
  for (int i = 1; i <= maxr; i++) {
    int t = i;
    while (t > 1) {
      tot[mindiv[t]] += vis[i];
      t /= prime[mindiv[t]];
    }
  }
}
inline bool check(long long x) {
  for (int i = 1; i <= cnt; i++) {
    long long t = 0, p = prime[i];
    while (x / p) {
      t += x / p;
      p *= prime[i];
    }
    if (t < tot[i]) return false;
  }
  return true;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    scanf("%d", &a[i]), vis[a[i]]++, r += a[i], maxr = max(maxr, a[i]);
  for (int i = maxr; i >= 1; i--) vis[i] += vis[i + 1];
  init();
  l = 1;
  while (l <= r) {
    long long mid = l + r >> 1;
    if (check(mid))
      r = mid - 1;
    else
      l = mid + 1;
  }
  printf("%I64d\n", l);
  return 0;
}