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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

#define si(n) scanf("%d",&n)
#define MAX 2003

int n,k,cnt[MAX][MAX];

int getSum(int x1,int y1,int x2,int y2)
{
    int xx= cnt[x2][y2]-cnt[x2][y1]-cnt[x1][y2]+cnt[x1][y1];
    return xx;
}

int solve(int x,int y)
{
    if(x<=k){
        return getSum(x,y,x+k,k)+getSum(1,1,x,y)
            + getSum(x+k,1,2*k,y);
    }
    return getSum(1,y,x-k,k)+getSum(x-k,1,x,y)
        + getSum(x,y,2*k,k);
}

int main()
{
   // freopen("input.txt","r",stdin);
    int i,j;
    si(n);si(k);
    for(i=1;i<=n;i++){
        int x,y;
        char ch;
        si(x);si(y);scanf(" %c",&ch);
        if(ch=='W')x+=k;
        x%=2*k;y%=2*k;
        if(y>=k){
            x+=k;y-=k;
        }
        x%=2*k;
        cnt[x+1][y+1]++;
        //cout<<x+1<<' '<<y+1<<' '<<cnt[x+1][y+1]<<endl;
    }
    //exit(0);
    for(i=1;i<=2*k;i++){
        for(j=1;j<=k;j++){
            cnt[i][j]+=cnt[i-1][j]+cnt[i][j-1]-cnt[i-1][j-1];
        }
    }

   // exit(0);
    int ans=0;
    for(i=0;i<2*k;i++){
        for(j=0;j<k;j++){
            int x=solve(i,j);
            //cout<<i<<' '<<j<<' '<<x<<endl;
            ans=max(ans,x);
        }
    }
    printf("%d\n",ans);

    return 0;
}