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<cstdlib>
#include<queue>
#include<set>
#include<vector>
#include<string>
#include<algorithm>
#include<stack>
#include<map>
#include<cstdio>
using namespace std;
#define rep(i,a) for(int i=0;i<a;i++)
#define rep2(i,a,b)for(int i=a;i<b;i++)
#define mp make_pair
#define pb push_back
#define P pair<int,int>
#define ll __int64
//#define ll long long

int n,k;
int x[111111],y[111111];

int rui[4001][4001];

int q(int x1,int y1,int x2,int y2){//半開区間での中のruiを返す
	
	if(x1>x2||y1>y2)return 0;
	int ret=0;
	ret=rui[y2][x2]-rui[y2][x1-1]-rui[y1-1][x2]+rui[y1-1][x1-1];

	return ret;
}

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



	rep(i,n){
		int ny=y[i],nx=x[i];
		rui[ny][nx]++;
	}


rep2(i,1,2*k+1){
		rep2(j,2,2*k+1)rui[i][j]+=rui[i][j-1];
	}
	
	rep2(i,1,2*k+1){
		rep2(j,2,2*k+1)rui[j][i]+=rui[j-1][i];
	}
	


	int ans=0;
	rep2(i,1,k+1){
		rep2(j,1,k+1){
			int tmp=q(i,j,i+k-1,j+k-1)+q(1,1,i-1,j-1)+q(i+k,1,2*k,j-1)+q(1,j+k,i-1,2*k)+q(i+k,j+k,2*k,2*k);
			ans=max(ans,max(n-tmp,tmp));
		}
	}
	
	cout<<ans<<endl;
	return 0;
}