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 N = 1e6 + 5;
const int Mx = 1e7;
const int mod = 1e9 + 7;
int n, k, a[N], lp[Mx + 5], b[Mx + 5];
long long ans = 1, cnt[Mx + 5];
vector<int> p;
long long calc(long long x, long long y) {
  long long ans = 0;
  while (x) {
    x /= y;
    ans += x;
  }
  return ans;
}
int main() {
  for (int i = 2; i <= Mx; i++) {
    if (lp[i] == 0) {
      p.push_back(i);
      for (int j = i; j <= Mx; j += i)
        if (!lp[j]) lp[j] = i;
    }
  }
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[a[i]]++;
  }
  for (int i = Mx; i; i--) {
    b[i] += b[i + 1];
    if (b[i] == 0) continue;
    int x = i;
    while (x > 1) {
      int y = lp[x];
      while (x % y == 0) {
        cnt[y] += b[i];
        x /= y;
      }
    }
  }
  for (auto i : p) {
    if (cnt[i] == 0) continue;
    long long l = 1, r = 1ll * i * cnt[i];
    long long cur = -1;
    while (l <= r) {
      long long mid = (l + r) / 2;
      if (calc(mid, i) >= cnt[i]) {
        cur = mid;
        r = mid - 1;
      } else
        l = mid + 1;
    }
    ans = max(ans, cur);
  }
  cout << ans;
  return 0;
}