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 <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <map>
#include <set>
#include <queue>

using namespace std;

#define lli long long int
#define REP(i,n) for(int i=0;i<n;i++)
#define REP2(i,n) for(int i=1;i<=n;i++)
#define DEBUG 0

lli dp[2010][2010];
lli s[4002][4002];

lli get(lli x,lli y,lli k){
	return s[x+k-1][y+k-1]+s[x-1][y-1]-s[x-1][y+k-1]-s[x+k-1][y-1];
}

int main(){

	lli n,k;
	cin>>n>>k;

	REP(i,2010)REP(j,2010)dp[i][j]=0;

	REP(i,n){
		lli x,y;
		char c;
		cin>>x>>y>>c;
		if(c=='W')y+=k;
		dp[x%(2*k)+1][y%(2*k)+1]++;
	}

	for(lli j=1;j<4*k+1;j++){
		for(lli i=1;i<4*k+1;i++){
			s[i][j] = s[i][j-1]+dp[(i-1)%(2*k)+1][(j-1)%(2*k)+1];
		}
	}

	for(lli j=1;j<4*k+1;j++){
		for(lli i=1;i<4*k+1;i++){
			s[i][j] += s[i-1][j];
		}
	}
	lli ans=0;

	for(lli i=1;i<2*k+1;i++){
		for(lli j=1;j<2*k+1;j++){
			ans = max(ans,get(i,j,k)+get(i+k,j+k,k));
		}
	}


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