Maximum Value(Codeforces_484B)

You are given a sequence a consisting of n integers. Find the maximum possible value of (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.

Sample test(s)

Input

3

3 4 5

Output

2

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>

using namespace std;

int main()
{
int n;
int a,num[2000100];
scanf("%d",&n);
int maxx=-99999;
for(int i=1;i<=n;i++)
{
scanf("%d",&a);
num[a]=a;
maxx=max(maxx,a);
}
for(int i=1;i<=maxx*2+1;i++)
{
if(num[i]!=i) num[i]=num[i-1];//这样处理一下更高效
}
int ans=0;
for(int i=2;i<maxx;i++)
{
if(num[i]==i)
{
for(int j=i*2;j<=maxx*2+1;j+=i)//枚举i的倍数
{
if(num[j-1]>i)//与i取余所能得到的最大值只会在比i的倍数小的数中的最大值中出现
ans=max(ans,num[j-1]%i);//（有点绕，譬如有四个数3,4,5,6,i=3，比i的2倍小的数有4,5,最大的是5,所以5%3是当前所得到的最大余数
}
}
}
printf("%d\n",ans);
return 0;
}