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 = 10000100;
const long long LIM = 1e14;
int criba[MAXN + 1];
void crearcriba() {
  int w[] = {4, 2, 4, 2, 4, 6, 2, 6};
  for (int p = 25; p <= MAXN; p += 10) criba[p] = 5;
  for (int p = 9; p <= MAXN; p += 6) criba[p] = 3;
  for (int p = 4; p <= MAXN; p += 2) criba[p] = 2;
  for (int p = 7, cur = 0; p * p <= MAXN; p += w[cur++ & 7])
    if (!criba[p]) {
      for (int j = p * p; j <= MAXN; j += (p << 1))
        if (!criba[j]) criba[j] = p;
    }
}
long long cant(long long p, long long n) {
  long long ans = 0, k = 1;
  while (k <= n / p) k *= p, ans += n / k;
  return ans;
}
int n, H[MAXN];
long long acum[MAXN], cnt;
map<int, long long> P;
int main() {
  crearcriba();
  while (cin >> n) {
    P.clear();
    cnt = 0;
    for (int i = (0); i < (int)(MAXN); i++) H[i] = 0, acum[i] = 0;
    for (int i = (0); i < (int)(n); i++) {
      int x;
      scanf("%d", &x);
      H[x]++;
    }
    for (int i = ((int)MAXN) - 1; i >= 0; i--) {
      if (i < 2) break;
      cnt += H[i];
      long long k = acum[i] + cnt;
      if (!k) continue;
      int p = criba[i];
      if (p) {
        acum[p] += k;
        acum[i / p] += k;
      } else
        P[i] = k;
    }
    long long ans = 1;
    for (auto it = P.begin(); it != P.end(); ++it) {
      long long p = it->first, e = it->second;
      long long l = 0, m, r = LIM;
      while (r - l > 1) {
        m = (l + r) / 2;
        if (cant(p, m) >= e)
          r = m;
        else
          l = m;
      }
      ans = max(ans, r);
    }
    printf("%lld\n", ans);
  }
  return 0;
}