C. Cellular Network

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given n points on the straight line — the positions (x-coordinates) of the cities and m points on the same line — the positions (x-coordinates) of the cellular towers. All towers work in the same way — they provide cellular network for all cities, which are located at the distance which is no more than r from this tower.

Your task is to find minimal r that each city has been provided by cellular network, i.e. for each city there is at least one cellular tower at the distance which is no more than r.

If r = 0 then a tower provides cellular network only for the point where it is located. One tower can provide cellular network for any number of cities, but all these cities must be at the distance which is no more than rfrom this tower.

Input

The first line contains two positive integers n and m (1 ≤ n, m ≤ 105) — the number of cities and the number of cellular towers.

The second line contains a sequence of n integers a1, a2, ..., an ( - 109 ≤ ai ≤ 109) — the coordinates of cities. It is allowed that there are any number of cities in the same point. All coordinates ai are given in non-decreasing order.

The third line contains a sequence of m integers b1, b2, ..., bm ( - 109 ≤ bj ≤ 109) — the coordinates of cellular towers. It is allowed that there are any number of towers in the same point. All coordinates bj are given in non-decreasing order.

Output

Print minimal r so that each city will be covered by cellular network.

Examples

input

3 2-2 2 4-3 0

output

4

input

5 31 5 10 14 174 11 15

output

3

 

题目链接：

让每一个城市都可以被供给，也就是说每一个发电站的距离向左右延伸后要把城市最小~最大范围全部覆盖到，想了一会儿感觉是找每一个城市的最近供给站的距离，然后取最大值作为标准，然后进行二分……，然后怎么找这个最近的供给站呢？还是二分……，找的时候注意边界就可以了

代码：

 

#include
     
      using namespace std;#define INF 0x3f3f3f3f#define MM(x,y) memset(x,y,sizeof(x))#define LC(x) (x<<1)#define RC(x) ((x<<1)+1)#define MID(x,y) ((x+y)>>1)typedef pair
      
        pii;typedef long long LL;const double PI=acos(-1.0);const int N=100010;int n,m;int c[N];int d[N];int vis[N];LL min_dis[N];inline LL ABS(const LL &a){	return a>0?a:-a;}int main(void){	int i,j,k;	while (~scanf("%d%d",&n,&m))	{		for (i=0; i
       
        =m)				l=m-1;			if(r>=m)				r=m-1;			if(r<0)				r=0;			min_dis[i]=min(ABS((LL)c[i]-d[l]),ABS((LL)c[i]-d[r]));			if(min_dis[i]>dx)				dx=min_dis[i];		}		int nm=n+m;		LL l=0,r=2LL*1e9,mid,ans;		while (l<=r)		{			mid=(l+r)>>1;			if(mid>=dx)			{				r=mid-1;				ans=mid;			}			else				l=mid+1;		}		printf("%I64d\n",ans);	}	return 0;}