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;
template <class T>
bool umin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool umax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
int suf[10000009];
long long all[10000009];
bool vis[10000009];
long long ok(long long x, long long c) {
  long long res = 0;
  while (x >= c) {
    res += x / c;
    x /= c;
  }
  return res;
}
int main() {
  int n, mx = 0;
  long long sum = 0;
  scanf("%d", &n);
  while (n--) {
    int x;
    scanf("%d", &x);
    suf[x]++;
    sum += x;
    umax(mx, x);
  }
  for (int i = mx; i >= 1; i--) suf[i] += suf[i + 1];
  for (int i = 2; i < 10000009; i++) {
    if (vis[i]) continue;
    for (int j = 1; j * i <= 10000000; j++) {
      if (j > 1) vis[i * j] = true;
      int cnt = 1;
      if (i == 2)
        cnt += __builtin_ctz(j);
      else {
        for (int k = j; k % i == 0; k /= i) cnt++;
      }
      all[i] += cnt * suf[j * i];
    }
    if (i > 2) i++;
  }
  long long ans = 1;
  for (int i = 2; i <= 10000000; i++) {
    if (vis[i]) continue;
    long long st = 0, en = sum;
    while (st + 1 < en) {
      long long mid = (st + en) / 2;
      if (ok(mid, i) >= all[i])
        en = mid;
      else
        st = mid;
    }
    if (ok(st, i) >= all[i])
      umax(ans, st);
    else
      umax(ans, en);
  }
  printf("%lld\n", ans);
  return 0;
}