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 rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

int main(){
	int n,k; cin>>n>>k;
	vector<vector<int>> B(2*k+1,vector<int>(2*k+1));
	vector<vector<int>> W(2*k+1,vector<int>(2*k+1));
	rep(i,n){
		int x,y;
		char c; cin>>x>>y>>c;
		if(c=='B') B[y%(2*k)+1][x%(2*k)+1]++;
		else       W[y%(2*k)+1][x%(2*k)+1]++;
	}
	rep(i,2*k) rep(j,2*k) {
		B[i+1][j+1]+=B[i+1][j]+B[i][j+1]-B[i][j];
		W[i+1][j+1]+=W[i+1][j]+W[i][j+1]-W[i][j];
	}

	int ans=0;
	rep(i,k+1) rep(j,k+1) {
		int bcnt=0,wcnt=0;
		bcnt+=B[k+i][k+j]-B[i][k+j]-B[k+i][j]+B[i][j];
		bcnt+=B[2*k][2*k]-B[k+i][2*k]-B[2*k][k+j]+B[k+i][k+j];
		bcnt+=B[i][j];
		bcnt+=B[i][2*k]-B[i][k+j];
		bcnt+=B[2*k][j]-B[k+i][j];
		wcnt+=W[k+i][2*k]-W[k+i][k+j]-W[i][2*k]+W[i][k+j];
		wcnt+=W[2*k][k+j]-W[k+i][k+j]-W[2*k][j]+W[k+i][j];
		wcnt+=W[i][k+j]-W[i][j];
		wcnt+=W[k+i][j]-W[i][j];
		ans=max({ans,bcnt+wcnt,n-bcnt-wcnt});
	}
	cout<<ans<<endl;

	return 0;
}
