Problem: 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:

cba927b2484fad94fb5ff7473e9aadef.png

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.

Examples

Input

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


Output

4


Input

2 1000
0 0 B
0 1 W


Output

2


Input

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


Output

4

Solution (in CPP):

Solution:
#include<cstdio>
#include<cstring>
const int MAXN = 100000;
const int MAXK = 1000;
int x[MAXN + 5], y[MAXN + 5], c[MAXN + 5];
int y1[10 * MAXK + 5];
int main() {
    int N, K, Max = 0, cnt = 0;
    scanf("%d%d",&N,&K);
    for(int i=1;i<=N;i++) {
        char ch;
        scanf("%d%d %c",&x[i],&y[i],&ch);
        x[i] = x[i] % (2*K);//降维 
        y[i] = y[i] % (2*K);
        if( ch == 'W' ) c[i] = 1;
        else c[i] = 0;
    }
    for(int i=0;i<=K;i++) 
    {
        cnt = 0;
        memset(y1,0,sizeof(y1));
        for(int j=1;j<=N;j++) 
        {
            int p = ( i<=x[j] && x[j] < i+K );
            int q = ( 0 <=y[j] && y[j] < K );
            //printf("%d %d",p,q); 
            if( p ^ q == c[j] )//通过位与来实现
            {
                cnt++;
                y1[y[j]] ++;
            }
            else 
            {
                y1[y[j]] --;
            }
        }
        for(int j=0;j<=K;j++) 
        {
            if( cnt > Max ) Max = cnt;
            if( N - cnt > Max ) Max = N - cnt;
            if( j != K )
                cnt = cnt - y1[j] - y1[K+j];
        }
    }
    printf("%d\n",Max);
}