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 = 10000007;
long long a[maxn], p[maxn], ro[maxn], maxx = 0, sum = 0, num = 0;
bool fl[maxn];
long long inline read() {
  long long x = 0, w = 0;
  char ch = 0;
  while (ch < '0' || ch > '9') w |= ch == '-', ch = getchar();
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
  return w ? -x : x;
}
void get_prime(long long m) {
  for (long long i = 2; i <= m; i++) {
    if (!fl[i]) p[++num] = i;
    for (long long j = 2; j * i <= m; j++) {
      fl[i * j] = 1;
    }
  }
}
int check(long long mid) {
  for (int i = 1; i <= num; i++) {
    long long cnt = 0;
    for (long long j = p[i]; j <= mid; j *= p[i]) cnt += mid / j;
    if (cnt < ro[i]) return 0;
  }
  return 1;
}
int main() {
  int n = read();
  for (int i = 1; i <= n; i++) {
    long long k = read();
    a[k]++;
    sum += k;
    maxx = max(maxx, k);
  }
  get_prime(maxx);
  for (int i = maxx; i >= 1; i--) {
    a[i] += a[i + 1];
  }
  for (int i = 1; i <= num; i++) {
    for (long long j = p[i]; j <= maxx; j = 1ll * j * p[i]) {
      for (long long k = j; k <= maxx; k += j) {
        ro[i] += a[k];
      }
    }
  }
  long long l = 1, r = sum, ans;
  while (l <= r) {
    long long mid = (l + r) >> 1;
    if (check(mid)) {
      ans = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  printf("%lld", ans);
  return 0;
}