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 long long MAXLL = (1LL << 61);
const int N = 10000001;
const int M = 670000;
bool f[N];
int m[N];
int p[M];
long long cnt[M];
int size, Max;
void GetPrime() {
  for (int i = 2; i * i <= N; i++) {
    if (f[i]) continue;
    for (int j = 2 * i; j < N; j += i) {
      f[j] = 1;
    }
  }
  for (int i = 2; i < N; i++) {
    if (!f[i]) p[size++] = i;
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0, a; i < n; i++) {
    scanf("%d", &a);
    m[a]++;
    Max = max(Max, a);
  }
  for (int i = N - 2; i >= 0; i--) {
    m[i] += m[i + 1];
  }
  long long ans = 1;
  for (int i = 2; i < N; i++) {
    if (f[i]) continue;
    for (int j = 2 * i; j < N; j += i) {
      f[j] = 1;
    }
    long long t = 0;
    for (long long p = i; p < N && p > 0; p *= i) {
      for (long long q = p; q < N; q += p) {
        t += m[q];
      }
    }
    if (t == 0) continue;
    long long l = ans, r = MAXLL;
    while (l <= r) {
      long long mid = (l + r) / 2;
      long long tmp = mid, count = 0;
      while (tmp) {
        tmp /= i;
        count += tmp;
      }
      if (count >= t) {
        ans = mid;
        r = mid - 1;
      } else
        l = mid + 1;
    }
  }
  printf("%I64d\n", ans);
  return 0;
}