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<vector>
#include<algorithm>
using namespace std;

int cal(int x1,int y1, int x2, int y2, vector<vector<int>> &grid){
	return grid[y2][x2]-grid[y2][x1]-grid[y1][x2]+grid[y1][x1];
}

int main(){
	int n,k; cin>>n>>k;
	int x,y;
	char c;
	vector<vector<int>> grid(2*k+5,vector<int>(2*k+5,0));
	for(int i=0;i<n;i++){
		cin >> x >> y >> c;
		if(c=='W') x+=k;
		x%=2*k;
		y%=2*k;
		++grid[y+1][x+1];
	}

	int k2=2*k;
	for(int i=1;i<=k2;i++){
		for(int j=1;j<=k2;j++){
			grid[i][j]+=grid[i][j-1];
		}
	}
	for(int i=1;i<=k2;i++){
		for(int j=1;j<=k2;j++){
			grid[i][j]+=grid[i-1][j];
		}
	}
	int ans=0;
	for(int y=1;y<=k2;y++){
		for(int x=1;x<=k2;x++){
			int x1=0,x2=max(0,x-k),x3=x,x4=min(k2,x+k),x5=k2;
			int y1=0,y2=max(0,y-k),y3=y,y4=min(k2,y+k),y5=k2;
			int cnt=cal(x1,y1,x2,y2,grid)+cal(x3,y1,x4,y2,grid)+
					cal(x2,y2,x3,y3,grid)+cal(x4,y2,x5,y3,grid)+
					cal(x1,y3,x2,y4,grid)+cal(x3,y3,x4,y4,grid)+
					cal(x2,y4,x3,y5,grid)+cal(x4,y4,x5,y5,grid);
			ans=max(ans,cnt);
		}
	}
	cout << ans << endl;
	return 0;
}