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;

typedef long long ll;
typedef vector<ll> vl;
typedef vector<vector<ll>> vvl;
typedef pair<ll, ll> P;
#define rep(i, n) for(ll i = 0; i < n; i++)
#define REP(i, a, b) for(ll i = a; i <= b; i++)
#define out(x) std::cout << x << '\n'
#define all(a) a.begin(), a.end()
#define pb push_back
#define sz(x) ((int)(x).size())
#define chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { ll g = gcd(a, b); return a / g * b; }
const int dx[4] = { 1, 0, -1,  0 };
const int dy[4] = { 0, 1,  0, -1 };
const int INF = (int)1e9;
const ll  inf = (ll)1e18;
const int mod = 1000000007;
const int MAX_N = 100010;
const double PI = acos(-1);

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	cout << fixed << setprecision(10);
	/*------------------------------*/

	int n, m;
	cin >> n >> m;
	vector<int> a(n), b(n);
	rep(i, n) cin >> a[i] >> b[i];
	vector<int> c(m), d(m);
	rep(i, m) cin >> c[i] >> d[i];

	rep(i, n) {
		int mi = INF;
		int num = -1;
		rep(j, m) {
			int dd = abs(a[i] - c[j]) + abs(b[i] - d[j]);
			if (mi > dd) {
				mi = dd;
				num = j + 1;
			}
		}
		out(num);
	}

	return 0;
}