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 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

4

input

5 3
1 5 10 14 17
4 11 15

output

3



题意：有n座城市，和m座信号塔，要你找出这些信号塔能覆盖所有城市的最短距离。

题解： 对信号塔b进行二分,找城市与信号塔的最小距离即可。（题目给的数据是严格非连续递增的）


AC代码：


#include<bits/stdc++.h>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<map>
#include<queue>
#include<set>
#include<stack>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define MP(x,y) make_pair(x,y)  
template <class T1, class T2>inline void getmax(T1 &a, T2 b) { if (b>a)a = b; }  
template <class T1, class T2>inline void getmin(T1 &a, T2 b) { if (b<a)a = b; }
int read()
{
  int v = 0, f = 1;
  char c =getchar();
  while( c < 48 || 57 < c ){
    if(c=='-') f = -1;
    c = getchar();
  }
  while(48 <= c && c <= 57) 
    v = v*10+c-48, c = getchar();
  return v*f;
}

int main()
{  
  //freopen("in.txt","r",stdin);
  int n,m;
  cin>>n>>m;
  int a[n+1],b[m+1];
  for(int i = 0; i < n; i++){
    cin>>a[i];
  }
  for (int i = 0; i < m; i++)
  {
    cin>>b[i];
  }
  int ans = 0 ;
  for(int i = 0; i<n; i++){
    int l = 0, r = m-1; //对信号塔b进行二分,找城市与信号塔的最小距离
    while(r-l > 1){
      int mid = (r+l)>>1;
      if(b[mid] >= a[i]){
        r = mid;
      }
      else{

        l=mid;
      }
    }
    int cnt = min(abs(a[i]-b[l]),abs(a[i]-b[r]));
    ans = max(ans, cnt);
  }
  cout<<ans;
  return 0;
}