Problem:
Score : 500 points

Problem Statement

AtCoDeer is thinking of painting an infinite two-dimensional grid in a checked pattern of side K.
Here, a checked pattern of side K is a pattern where each square is painted black or white so that each connected component of each color is a K × K square.
Below is an example of a checked pattern of side 3:

AtCoDeer has N desires.
The i-th desire is represented by x_i, y_i and c_i.
If c_i is B, it means that he wants to paint the square (x_i,y_i) black; if c_i is W, he wants to paint the square (x_i,y_i) white.
At most how many desires can he satisfy at the same time?

Constraints

1 ≤ N ≤ 10^5

1 ≤ K ≤ 1000

0 ≤ x_i ≤ 10^9

0 ≤ y_i ≤ 10^9

If i ≠ j, then (x_i,y_i) ≠ (x_j,y_j).

c_i is B or W.

N, K, x_i and y_i are integers.

Input

Input is given from Standard Input in the following format:

N K
x_1 y_1 c_1
x_2 y_2 c_2
:
x_N y_N c_N

Output

Print the maximum number of desires that can be satisfied at the same time.

Sample Input 1

4 3
0 1 W
1 2 W
5 3 B
5 4 B

Sample Output 1

4

He can satisfy all his desires by painting as shown in the example above.

Sample Input 2

2 1000
0 0 B
0 1 W

Sample Output 2

2

Sample Input 3

6 2
1 2 B
2 1 W
2 2 B
1 0 B
0 6 W
4 5 W

Sample Output 3

4

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

#include<bits/stdc++.h> 
using namespace std;  
const int K = 3005;  
int x[K][K];  
int n,k;  
int getsize(int x1,int y1,int x2,int y2){  
    return x[x2][y2]-x[x2][y1]-x[x1][y2]+x[x1][y1];  
}  
int main(){  
    scanf("%d%d",&n,&k);  
    int a,b,i,j;  
    char c;  
    for(i = 1; i <= n; i++){  
        scanf("%d%d %c",&a,&b,&c);  
        a %= (2*k);  
        b %= (2*k);  
        if(c == 'W')  
            b += k;  
        x[a%(2*k)+1][b%(2*k)+1]++;  
    }  
    for(i = 1; i <= 2*k; i++){  
        for(j = 1; j <= 2*k; j++){  
            x[i][j] = x[i][j]+x[i-1][j]+x[i][j-1]-x[i-1][j-1];  
        }  
    }  
    int ans = 0,sum;  
    for(i = 1; i <= k; i++){  
        for(j = 1; j <= k; j++){  
            sum = getsize(0,0,i,j);  
            sum += getsize(i,j,k+i,k+j);  
            sum += getsize(k+i,k+j,2*k,2*k);  
            sum += getsize(k+i,0,2*k,j);  
            sum += getsize(0,k+j,i,2*k);  
            sum = max(n-sum,sum);  
            ans = max(ans,sum);  
        }  
    }  
    printf("%d\n",ans);  
    return 0;  
}