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 = 10000005;
int limit, n, tn;
long long sum;
int a[MAXN], p[MAXN], DIV[MAXN], b[MAXN];
long long tot[MAXN];
bool can[MAXN];
void Prepare() {
  memset(can, 0, sizeof(can));
  for (int i = 2; i <= limit; ++i) {
    if (!can[i]) p[++tn] = i, DIV[i] = tn;
    for (int j = 1; j <= tn && i * p[j] <= limit; ++j) {
      can[i * p[j]] = 1;
      DIV[i * p[j]] = j;
      if (i % p[j] == 0) break;
    }
  }
  for (int i = 1; i <= limit; ++i) {
    int t = i;
    while (t > 1) {
      tot[DIV[t]] += b[i];
      t /= p[DIV[t]];
    }
  }
}
bool Judge(long long x) {
  for (int i = 1; i <= tn; ++i) {
    long long t = 0, now = p[i];
    while (x / now) {
      t += x / now;
      now = now * p[i];
    }
    if (t < tot[i]) return 0;
  }
  return 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    limit = limit < a[i] ? a[i] : limit;
    sum += a[i], ++b[a[i]];
  }
  for (int i = limit; i >= 1; --i) b[i] += b[i + 1];
  Prepare();
  long long l = 1, r = sum, mid, ans;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (Judge(mid))
      r = mid - 1, ans = mid;
    else
      l = mid + 1;
  }
  cout << ans << endl;
  return 0;
}