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>
using namespace std;

int main(){
	int N,K,ans=0;
	cin >> N >> K;
	vector<int> x(N);
	vector<int> y(N);
	
	for(int i=0;i<N;i++){
		int X,Y;
		char C;
		cin >> X >> Y >> C;
		if(C=='W')Y+=K;
		X%=2*K;
		Y%=2*K;
		x[i]=X;
		y[i]=Y;
	}
	
	int board[2*K][2*K];
	int sum[2*K+1][2*K+1];
	for(int i=0;i<2*K;i++)for(int j=0;j<2*K;j++)board[i][j]=0;
	for(int i=0;i<2*K+1;i++)for(int j=0;j<2*K+1;j++)sum[i][j]=0;
	for(int i=0;i<N;i++)board[x[i]][y[i]]++;
	for(int i=0;i<2*K;i++)for(int j=0;j<2*K;j++){
		sum[i+1][j+1]=sum[i+1][j]+sum[i][j+1]-sum[i][j]+board[i][j];
	}
	
	for(int i=0;i<K;i++)for(int j=0;j<K;j++){
		int count=0;
		
		count+=sum[K+i][K+j]-sum[i][K+j]-sum[K+i][j]+sum[i][j];
		count+=sum[i][j];
		count+=sum[i][2*K]-sum[i][K+j];
		count+=sum[2*K][j]-sum[K+i][j];
		count+=sum[2*K][2*K]-sum[2*K][K+j]-sum[K+i][2*K]+sum[K+i][K+j];
		
		count=max(count,N-count);
		ans=max(ans,count);
	}
	cout << ans << endl;
	return 0;
}