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 <stdio.h>
#include <iostream>
#include <vector>
#include <algorithm>
#include <math.h>
 
#define FOR(i,to) for( int i=0;i<(to);i++)
#define FORR(i,from) for( int i=from;i>-1;i--)

using namespace std;

int N,K;
int s[4002][4002];
int w[4002][4002];

int calc(int x,int y){
	return w[x][y]-w[x-K][y]-w[x][y+K]+w[x-K][y+K];
}

int main(){
	cin >> N >> K;

	FOR(i,N){
		int x,y;
		char c;
		cin >> x >> y >> c;
		
		if(c=='W') s[(x+K)%(2*K)+K][y%(2*K)+K]+=1;
		else s[x%(2*K)+K][y%(2*K)+K]+=1; 
	}


	FOR(y,4*K) FOR(x,4*K){
		if(x==0) w[x][y]=s[x][y];
		else	  w[x][y]=w[x-1][y]+s[x][y];
	}

	FOR(x,4*K) FORR(y,4*K-1){
		if(y==4*K-1) w[x][y]=w[x][y];
		else	     w[x][y]=w[x][y+1]+w[x][y];
	}


	int ans=0;
	int tmp=0;

	for(int y=K; y<2*K; y++){
		for(int x=2*K; x<3*K; x++){
			tmp=calc(x,y)+calc(x-K,y-K)+calc(x+K,y-K)+calc(x+K,y+K)+calc(x-K,y+K);
			ans=max( ans, max(N-tmp,tmp) );
		}
	}

	cout << ans <<  endl;
	return 0;
}