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<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
const int MR=1e3+10;

typedef long long LL;

int f[MR][MR][2];
int main() {
	int n,K,x,y;
	int rx,ry,rc;
	char c;
	scanf("%d%d",&n,&K);
	for(int i=0;i<n;i++){
		scanf("%d%d %c",&x,&y,&c);
		rx=(x%K+K)%K;
		ry=(y%K+K)%K;
		if((x-rx+y-ry)/K%2){
			c='W'+'B'-c;
		}
		if(c=='W') rc=0;
		else rc=1;
		f[rx][ry][rc]+=1;f[rx][K][rc]-=1;f[K][ry][rc]-=1;f[K][K][rc]+=1;
		f[0][0][rc]+=1;f[rx][0][rc]-=1;f[0][ry][rc]-=1;f[rx][ry][rc]+=1;
		rc=1-rc;
		f[rx][0][rc]+=1;f[K][0][rc]-=1;f[rx][ry][rc]-=1;f[K][ry][rc]+=1;
		f[0][ry][rc]+=1;f[0][K][rc]-=1;f[rx][ry][rc]-=1;f[rx][K][rc]+=1;
	} 
	int ans=-1;
	for(int l=0;l<2;l++)
	for(int i=0;i<K;i++){
		for(int j=0;j<K;j++){
			if(i>0 && j>0){
				f[i][j][l]+=f[i-1][j][l]+f[i][j-1][l]-f[i-1][j-1][l];
			}
			else if(i==0 && j>0){
				f[i][j][l]+=f[i][j-1][l];
			}
			else if(i>0 && j==0){
				f[i][j][l]+=f[i-1][j][l];
			}
			ans=max(ans,f[i][j][l]);
		}
	}

	printf("%d\n",ans);
    return 0;
}