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>
using namespace std;
#define FOR(i,k,n) for(int i = (int)(k); i < (int)(n); i++)
#define REP(i,n) FOR(i,0,n)
#define ALL(a) a.begin(), a.end()
#define MS(m,v) memset(m,v,sizeof(m))
typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<string> vs;
typedef pair<int, int> pii;
const int MOD = 1e9 + 7;
template<class T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template<class T> T &chmax(T &a, const T &b) { return a = max(a, b); }
template<class T>
istream& operator >> (istream& is, vector<T>& v)
{
	for (auto &i : v) is >> i;
	return is;
}
template<class T>
ostream& operator<<(ostream& os, vector<T>& v)
{
	const string delimiter = "\n";
	REP(i, v.size())
	{
		os << v[i];
		if (i != v.size() - 1) os << delimiter;
	}
	return os;
}
/*--------------------template--------------------*/

int main()
{
	cin.sync_with_stdio(false); cout << fixed << setprecision(10);
	int n, m;
	cin >> n >> m;
	vector<pii> v, w;
	REP(i, n)
	{
		int a, b; cin >> a >> b;
		v.emplace_back(a, b);
	}
	REP(i, m)
	{
		int a, b; cin >> a >> b;
		w.emplace_back(a, b);
	}
	REP(i, n)
	{
		int res = -1, dist = 1e9;
		REP(j, m)
		{
			int d = abs(v[i].first - w[j].first) + abs(v[i].second - w[j].second);
			if (d < dist)
			{
				res = j + 1;
				dist = d;
			}
		}
		cout << res << endl;
	}
	return 0;
}