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
#include<bits/stdc++.h>
#define INF 0x3f3f3f3f
const int N = 1000+5;
using namespace std;
int G[N*2][N*2];
int getNum(int x1,int y1,int x2,int y2){
    return G[x2][y2]-G[x1][y2]-G[x2][y1]+G[x1][y1];
}
int main(){
    int n,k;
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++){
        int x,y;
        char c;
        scanf("%d %d %c",&x,&y,&c);
        if(c=='W')
            x+=k;
        x=x%(k*2);
        y=y%(k*2);
        G[x+1][y+1]++;
    }
    for(int i=1;i<=2*k;i++)
        for(int j=1;j<=2*k;j++)
            G[i][j]+=G[i-1][j]+G[i][j-1]-G[i-1][j-1];
    int res=-INF;
    for(int i=1;i<=k;i++){
        for(int j=1;j<=k;j++){
            int temp1=getNum(0,0,i,j);
            int temp2=getNum(0,j+k,i,2*k);
            int temp3=getNum(i+k,0,2*k,j);
            int temp4=getNum(i+k,j+k,2*k,2*k);
            int temp5=getNum(i,j,i+k,j+k);
            int sum=temp1+temp2+temp3+temp4+temp5;
            res=max(res,max(sum,n-sum));
        }
    }
    printf("%d\n",res);
    return 0;
}