#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 7;
int ar[N];
int dd[N];
int n, mx, rs, x;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    ar[x] = x;
    mx = max(mx, x);
  }
  for (int i = 1; i <= 2 * mx; i++)
    if (ar[i] == 0) ar[i] = ar[i - 1];
  for (int i = 1; i <= 2 * mx; i++) {
    if (ar[i] != i) continue;
    for (int j = 2 * i - 1; j <= 2 * mx; j += i) rs = max(rs, ar[j] % i);
  }
  cout << rs << endl;
  return 0;
}

### Prompt

Please formulate a CPP solution to the following problem:

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;
const int N = 2e6 + 7;
int ar[N];
int dd[N];
int n, mx, rs, x;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    ar[x] = x;
    mx = max(mx, x);
  }
  for (int i = 1; i <= 2 * mx; i++)
    if (ar[i] == 0) ar[i] = ar[i - 1];
  for (int i = 1; i <= 2 * mx; i++) {
    if (ar[i] != i) continue;
    for (int j = 2 * i - 1; j <= 2 * mx; j += i) rs = max(rs, ar[j] % i);
  }
  cout << rs << endl;
  return 0;
}

```