#include <bits/stdc++.h>
using namespace std;
int a[212345];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  sort(a, a + n);
  int ans = 0;
  int last = 1;
  for (int i = 0; i < n; i++) {
    if (a[i] == last) continue;
    last = a[i];
    int l = 0;
    int x = a[i];
    for (int j = 2;; j++) {
      x += a[i];
      int p = (lower_bound(a + l, a + n, x) - a) - 1;
      if (p < 0) continue;
      ans = max(ans, a[p] % a[i]);
      if (x > a[n - 1]) break;
    }
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

You are given a sequence a consisting of n integers. Find the maximum possible value of <image> (integer remainder of ai divided by aj), where 1 ≤ i, j ≤ n and ai ≥ aj.

Input

The first line contains integer n — the length of the sequence (1 ≤ n ≤ 2·105). 

The second line contains n space-separated integers ai (1 ≤ ai ≤ 106).

Output

Print the answer to the problem.

Examples

Input

3
3 4 5


Output

2

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int a[212345];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  sort(a, a + n);
  int ans = 0;
  int last = 1;
  for (int i = 0; i < n; i++) {
    if (a[i] == last) continue;
    last = a[i];
    int l = 0;
    int x = a[i];
    for (int j = 2;; j++) {
      x += a[i];
      int p = (lower_bound(a + l, a + n, x) - a) - 1;
      if (p < 0) continue;
      ans = max(ans, a[p] % a[i]);
      if (x > a[n - 1]) break;
    }
  }
  printf("%d\n", ans);
  return 0;
}

```