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;
int prime[10000005];
void sieve(int n) {
  for (int i = 1; i <= n; i++) {
    prime[i] = i;
    if ((i & 1) == 0) prime[i] = 2;
  }
  for (int i = 3, lim = sqrt(n); i <= lim; i += 2)
    if (prime[i] == i)
      for (int j = i * i; j <= n; j += i) prime[j] = i;
}
int K, a;
long long e[10000005];
int qf;
pair<int, long long> f[10000005];
int main() {
  memset(e, 0, sizeof(e));
  long long lo = 0, hi = 0;
  scanf("%d", &K);
  for (int i = 0; i < K; i++) {
    cin >> a;
    hi += a;
    lo = max(lo, (long long)a);
    e[a]++;
  }
  sieve(lo);
  for (int i = lo; i > 1; i--) e[i] += e[i + 1];
  for (int i = lo; i > 1; i--) {
    if (e[i] > 0 && prime[i] != i) {
      int f = prime[i];
      int cf = i / f;
      e[f] += e[i];
      e[cf] += e[i];
      e[i] = 0;
    }
  }
  qf = 0;
  for (int i = 2; i <= lo; i++)
    if (e[i]) f[qf++] = make_pair(i, e[i]);
  while (lo < hi) {
    long long mid = (lo + hi) >> 1;
    bool flag = 1;
    for (int i = 0; i < qf; i++) {
      long long cnt = 0, x = f[i].first;
      while (x <= mid) {
        cnt += mid / x;
        x *= f[i].first;
      }
      if (cnt < f[i].second) {
        flag = 0;
        break;
      }
    }
    if (flag)
      hi = mid;
    else
      lo = mid + 1;
  }
  cout << lo << endl;
}