Educational Codeforces Round 15 C 二分

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 r from 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

input

5 3
1 5 10 14 17
4 11 15

output

题意：给你n个城市的坐标和m个塔的坐标 (都在x轴上,非递减的给出) 塔的工作范围为半径为r的圆

问最小的r使得所有的城市都在塔的工作范围内。

题解：枚举每个城市二分得到最近的塔的位置取距离差值的max

 1 /******************************
 2 code by drizzle
 3 blog: www.cnblogs.com/hsd-/
 4 ^ ^    ^ ^
 5  O      O
 6 ******************************/
 7 //#include<bits/stdc++.h>
 8 #include<iostream>
 9 #include<cstring>
10 #include<cstdio>
11 #include<map>
12 #include<algorithm>
13 #include<queue>
14 #define ll __int64
15 using namespace std;
16 int n,m;
17 int a[100005];
18 int b[100005];
19 int main()
20 {
21     scanf("%d %d",&n,&m);
22     for(int i=0; i<n; i++)
23         scanf("%d",&a[i]);
24     for(int i=0; i<m; i++)
25         scanf("%d",&b[i]);
26     int ans=0;
27     for(int i=0; i<n; i++)
28     {
29         int pos=lower_bound(b,b+m,a[i])-b;
30         if(pos==0)
31             ans=max(ans,b[pos]-a[i]);
32         else if(pos==m)
33             ans=max(ans,a[i]-b[pos-1]);
34         else
35             ans=max(ans,min(b[pos]-a[i],a[i]-b[pos-1]));
36     }
37     cout<<ans<<endl;
38     return 0;
39 }