Solve in Python:
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-----
2 2
2 0
0 0
-1 0
1 0

-----Sample Output-----
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.
n, m = list(map(int,input().split()))
s_x_plus_y = []
s_x_minus_y = []
p_x_plus_y = []
p_x_minus_y = []

for student in range(n):
    x, y = list(map(int,input().split()))
    s_x_plus_y.append(x + y)
    s_x_minus_y.append(x - y)

for i in range(m):
    x, y = list(map(int,input().split()))
    p_x_plus_y.append(x + y)
    p_x_minus_y.append(x - y)


for i in range(n):
    p_dist = []
    for j in range(m):
        p1 = abs(s_x_plus_y[i] - p_x_plus_y[j])
        p2 = abs(s_x_minus_y[i] - p_x_minus_y[j])
        d = max(p1, p2)
        p_dist.append(d)
    print((p_dist.index(min(p_dist))+1))