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;
double PI = acos(-1);
int const N = 1e6 + 10, M = 1e7 + 1, oo = 1e9;
const int mod = oo + 7;
const long long OO = 3e18;
int n, aa[N], pr[M];
long long cnt[M];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  for (int i = 2; i * i < M; i++) {
    if (pr[i]) continue;
    pr[i] = i;
    for (int j = i; j < M; j += i) pr[j] = i;
  }
  for (int i = 2; i < M; i++)
    if (pr[i] == 0) pr[i] = i;
  cin >> n;
  for (int i = 0, a; i < n; i++) {
    cin >> a;
    cnt[a]++;
  }
  for (int i = M - 2; i > 0; i--) cnt[i] += cnt[i + 1];
  for (int i = M - 1; i > 0; i--) {
    if (cnt[i] == 0 || pr[i] == i) continue;
    int y = i;
    while (y > 1) {
      int r = pr[y], c = 0;
      while (y % r == 0) y /= r, c++;
      cnt[r] += c * 1LL * cnt[i];
    }
  }
  long long an = 1;
  for (int i = 2; i < M; i++) {
    if (pr[i] != i || cnt[i] == 0) continue;
    long long l = 1, r = OO, mid, res = 0;
    while (l <= r) {
      mid = (l + r) >> 1;
      long long op = mid, c = 0;
      while (op) {
        op /= i;
        c += op;
      }
      if (c >= cnt[i]) {
        res = mid;
        r = mid - 1;
      } else
        l = mid + 1;
    }
    an = max(an, res);
  }
  cout << an;
  return 0;
}