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 MAXN = 10000001;
long long num, p[MAXN], a[MAXN], a_[MAXN];
bool isp[MAXN];
void init(int n) {
  memset(isp, true, sizeof(isp));
  isp[1] = false;
  for (int i = 2; i <= n; i++) {
    if (isp[i]) p[++num] = i;
    for (int j = 1; j <= num && i * p[j] <= n; j++) {
      isp[i * p[j]] = false;
      if (!i % p[j]) break;
    }
  }
}
bool check(long long mid) {
  for (int i = 1; i <= num; i++) {
    long long now = p[i], tmp = 0;
    while (mid >= now) {
      tmp = tmp + mid / now;
      now = now * p[i];
    }
    if (tmp < a[i]) return false;
  }
  return true;
}
int main() {
  int n;
  long long l = 0, r = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    long long x;
    scanf("%lld", &x);
    a_[x]++;
    l = max(l, x);
    r = r + x;
  }
  init(l);
  for (int i = l; i >= 1; i--) a_[i] = a_[i] + a_[i + 1];
  for (int i = 1; i <= num; i++) {
    for (long long j = p[i]; j <= l; j = j * p[i]) {
      for (int k = j; k <= l; k = k + j) a[i] = a[i] + a_[k];
    }
  }
  long long ans = r;
  while (l <= r) {
    long long mid = (l + r) / 2;
    if (check(mid)) {
      r = mid - 1;
      ans = mid;
    } else
      l = mid + 1;
  }
  printf("%lld", ans);
}