Problem:
Score : 200 points

Problem Statement

There are N students and M checkpoints on the xy-plane.

The coordinates of the i-th student (1 \leq i \leq N) is (a_i,b_i), and the coordinates of the checkpoint numbered j (1 \leq j \leq M) is (c_j,d_j).

When the teacher gives a signal, each student has to go to the nearest checkpoint measured in Manhattan distance.

The Manhattan distance between two points (x_1,y_1) and (x_2,y_2) is |x_1-x_2|+|y_1-y_2|.

Here, |x| denotes the absolute value of x.

If there are multiple nearest checkpoints for a student, he/she will select the checkpoint with the smallest index.

Which checkpoint will each student go to?

Constraints

1 \leq N,M \leq 50

-10^8 \leq a_i,b_i,c_j,d_j \leq 10^8

All input values are integers.

Input

The input is given from Standard Input in the following format:

N M
a_1 b_1
:
a_N b_N
c_1 d_1
:
c_M d_M

Output

Print N lines.

The i-th line (1 \leq i \leq N) should contain the index of the checkpoint for the i-th student to go.

Sample Input 1

2 2
2 0
0 0
-1 0
1 0

Sample Output 1

2
1

The Manhattan distance between the first student and each checkpoint is:

For checkpoint 1: |2-(-1)|+|0-0|=3

For checkpoint 2: |2-1|+|0-0|=1

The nearest checkpoint is checkpoint 2. Thus, the first line in the output should contain 2.

The Manhattan distance between the second student and each checkpoint is:

For checkpoint 1: |0-(-1)|+|0-0|=1

For checkpoint 2: |0-1|+|0-0|=1

When there are multiple nearest checkpoints, the student will go to the checkpoint with the smallest index. Thus, the second line in the output should contain 1.

Sample Input 2

3 4
10 10
-10 -10
3 3
1 2
2 3
3 5
3 5

Sample Output 2

3
1
2

There can be multiple checkpoints at the same coordinates.

Sample Input 3

5 5
-100000000 -100000000
-100000000 100000000
100000000 -100000000
100000000 100000000
0 0
0 0
100000000 100000000
100000000 -100000000
-100000000 100000000
-100000000 -100000000

Sample Output 3

5
4
3
2
1

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

typedef long long ll;
typedef long double ld;
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>Set;
typedef tree<int,int,less<int>,rb_tree_tag,tree_order_statistics_node_update> Map;

#define ipow(a,b)+0.5 (int)pow(a,b)
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define fi first
#define se second
#define pll pair<ll,ll>
#define ook order_of_key
#define findby find_by_order
#define pii pair<int,int>

#define rep(i,n) for(int i=0;i<n;i++)
#define reps(i,a,b) for(int i=a;i<=b;i++)

template<typename X> inline X abs(const X& a) { return (a < 0 ? -a : a); }
template<typename X> inline X sqr(const X& a) { return (a * a); }

const ld EPS = ld(1e-9);
const ld PI = ld(3.141592653589);

struct comp
{
    bool operator()(pii a, pii b)
    {
        if(a.fi<b.fi)
        return true;
        else if (a.fi>b.fi)
        return false;
        else
        {
             if(a.se<b.se)
             return true;
             else
             return false;
        }
    }
};

map<int,pii>check;
vector<pii>store;
vector<int>res;

class Task
{
public:
    void solve(istream& in,ostream& out)
    {
          int n,m;
          in>>n>>m;

          int a,b;

          rep(i,n)
          {
                in>>a>>b;
                store.pb(mp(a,b));
          }

          rep(i,m)
          {
                in>>a>>b;
                check.insert(mp(i+1,mp(a,b)));
          }

          for(auto j=store.begin();j!=store.end();j++)
          {
                pii A=*j;

                ll mini=mod;
                ll store=0;
                for(auto h=check.begin();h!=check.end();h++)
                {
                      pii B=(*h).se;

                      if(dist(A,B)<mini)
                      {
                           mini=dist(A,B);
                           store=(*h).fi;
                      }
                }

                res.pb(store);
          }

          rep(i,n)
          {
                out<<res[i]<<endl;
          }
    }

    int dist(pii a, pii b)
    {
          return abs(a.fi-b.fi)+abs(a.se-b.se);
    }
};


int main()
{
    ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	Task solver;
	std::istream& in(std::cin);
	std::ostream& out(std::cout);
	solver.solve(in,out);
	out.flush();

    return 0;
}