Problem Description

Avin is observing the cars at a crossroads. He finds that there are n cars running in the east-west direction with the i-th car passing the intersection at time ai . There are another m cars running in the north-south direction with the i-th car passing the intersection at time bi . If two cars passing the intersections at the same time, a traffic crash occurs. In order to achieve world peace and harmony, all the cars running in the north-south direction wait the same amount of integral time so that no two cars bump. You are asked the minimum waiting time.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 1, 000). The second line contains n distinct integers ai (1 ≤ ai ≤ 1, 000). The third line contains m distinct integers bi (1 ≤ bi ≤ 1, 000).

Output

Print a non-negative integer denoting the minimum waiting time.

Sample Input

1 1

1

1

1  2

2

1 3

Sample Output

1

0

这道题目数据范围很小，直接暴力枚举就可以了，我们枚举等待时间，让南北方向的车每个都加上这等待时间，如果每个车都不跟东西方向的车时间重合，则输出答案

下面附上AC代码

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <queue>
using namespace std;
typedef long long ll;
ll a[100010];
ll b[100010];
int main()
{
    std::ios::sync_with_stdio(false);
    cin.tie(0),cout.tie(0);
    ll n,m;
    while(cin>>n>>m)
    {
        ll k;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(ll i=1;i<=n;i++)
        {
            cin>>k;
            a[k]=1;
        }
        for(ll i=1;i<=m;i++)
        {
            cin>>b[i];
        }
        for(ll i=0;;i++)
        {
            int flag=0;
            for(ll j=1;j<=m;j++)
            {
                if(a[b[j]+i]==1)
                {
                    flag=1;
                    break;
                }
            }
            if(flag==0)
            {
                cout<<i<<endl;
                break;
            }
        }
    }
    return 0;
}