CodeForces

个人心得：周测题目，一题没出，难受得一批。这个题目做了一个半小时还是无限WR，虽然考虑到了二分答案这个点上面了，

奈何二分比较差就想用自己的优化，虽然卡在了a=k*b+c，这里但是后面结束了这样解决还是超时了，看了一下网上的hash，思想一样

但是却优化了很多，服气

题目：

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

Input

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

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

Output

Print the answer to the problem.

Example

Input

3
3 4 5

Output

2

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<iomanip>
#include<algorithm>
using namespace std;
#define inf 1<<29
#define infa 2000000+10
#define nu 50005
int n,m;
int book[infa];
int main()
{
    scanf("%d",&n);
    memset(book,-1,sizeof(book));
    int maxnum;
    for(int i=0;i<n;i++){
       scanf("%d",&m);
        book[m]=m;
    }
    for(int i=1;i<infa; i++)
        if(book[i]!=i)
        {
            book[i]=book[i-1];
        }
        int sum=0;
        for(int i=1;i<infa;i++)
        {
            if(book[i]==i)
                {
                    for(int j=i*2-1;j<infa;j+=i)
                        sum=max(sum,book[j]%i);
                }
        }
           cout<<sum<<endl;
    return 0;
}