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)

#pragma region
#include <bits/stdc++.h>
using namespace std;
inline void ci(void){
  return;
}
template <typename First, typename... Rest>
void ci(First& first, Rest&... rest){
  cin >> first;
  ci(rest...);
  return;
}
inline void co(void){
  cout << "\n";
  return;
}
template <typename First, typename... Rest>
void co(First first, Rest... rest){
  cout << first << " ";
  co(rest...);
  return;
}
#define ll long long
#define ld long double
#define st string
#define dl double
#define wh while
#define mp make_pair
#define mt make_tuple
#define pb push_back
#define tp tuple
#define ft first
#define sc second
#define ve vector
#define tos to_string
#define len length
#define rt return 0
#define br break
#define con continue
#define lcm(n, m) (n * m / gcd(n, m))
#define np next_permutation
#define repi(i, m, n) for(int i=(int)(m);i<(int)(n);i++)
#define rep(i, n) repi(i, 0, n)
#define repill(i, m, n) for(ll i=(ll)(m);i<(ll)(n);i++)
#define repll(i, n) repill(i, 0, n)
#define revi(i, m, n) for(int i=(int)(m);i>(int)(n);i--)
#define rev(i, n) revi(i, 0, n)
#define revill(i, m, n) for(ll i=(ll)(m);i>(ll)(n);i--)
#define revll(i, n) revill(i, 0, n)
#define sz(v) ((int)(v).size()) 
#define each(i, n) for(auto&& i: n)
#define all(n) (n.begin(), n.end())
#define allr(n) (n.rbegin(), n.rend())
#define sp(n) cout << fixed << setprecision(n)
#pragma endregion
// sqrt ルート, int 2*10^9, gcd　最大公約数
// __builtin_popcount bitの個数

int main() {
  int N, M;
  ci(N, M);
  ve<int> a(N), b(N), c(M), d(M);
  rep(i, N){
    ci(a.at(i), b.at(i));
  } 
  rep(i, M){
    ci(c.at(i), d.at(i));
  }
  ve<int> A(N, 0);
  rep(i, N){
    ll cnt = 2 * pow(10, 16);
    rep(j, M){
      ll x = abs(a.at(i) - c.at(j)) + abs(b.at(i) - d.at(j));
      if(cnt > x){
        cnt = x;
        A.at(i) = j;
      }
    }
  }
  rep(i, N){
    co(A.at(i) + 1);
  }
}