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: Runtime Error)

#include <stdio.h>
#include <stdlib.h>

int a[1001][2001]={0};

int main()
{
    int N, K;
    int x, y;
    char c;
    scanf("%d %d", &N, &K);
    int i,j;
    for(i=0;i<N;i++) {
        scanf("%d %d %c", &x, &y, &c);
        y%=2*K;
        if(c=='W')x+=K;
        if(y>=K){
            x+=K;
            y-=K;
        }
        a[y+1][x%(2*K)+1]++;
    }

    for(i=1;i<=K;i++){
        for(j=1;j<=2*K;j++){
            a[i+1][j]+=a[i][j];
        }
    }
    for(i=1;i<=K;i++){
        for(j=1;j<=2*K;j++){
            a[i][j+1]+=a[i][j];
        }
    }

    int ans=0;
    int tmp;
    for(i=1;i<=K;i++){
        for(j=1;j<=2*K;j++){
            if(j<=K+1){
                tmp=a[K][j+K-1]-a[K][j-1]-a[i-1][j+K-1]+a[i-1][j-1];
            }
            else{
                tmp=a[K][2*K]-a[K][j-1]-a[i-1][2*K-1]+a[i-1][j-1];
                tmp+=a[K][j-K-1]-a[K][0]-a[i-1][j-K-1]+a[i-1][0];
            }
            if(i>1){
                if(j>1&&j<=K+1){
                    tmp+=a[i-1][j-1]-a[i-1][0]-a[0][j-1]+a[0][0];
                }
                if(j>1&&j>K+1){
                    tmp+=a[i-1][j-1]-a[i-1][j-K-1]-a[0][j-1]+a[0][j-K-1];
                }
                if(j<=K){
                    tmp+=a[i-1][2*K]-a[i-1][j+K-1]-a[0][2*K]+a[0][j+K-1];
                }
            }
            if(ans<tmp)ans=tmp;
        }
    }
    printf("%d\n",ans);
    return 0;
}