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 <bits/stdc++.h>
 
using namespace std;
 
typedef long long LL;
typedef pair<int , int> P2;
typedef pair<pair<int , int> , int> P3;
typedef pair<pair<int , int> , pair<int , int> > P4;
#define PB(a) push_back(a)
#define MP(a , b) make_pair((a) , (b))
#define M3P(a , b , c) make_pair(make_pair((a) , (b)) , (c))
#define M4P(a , b , c , d) make_pair(make_pair((a) , (b)) , make_pair((c) , (d)))
#define repp(i,a,b) for(int i = (int)(a) ; i < (int)(b) ; ++i)
#define repm(i,a,b) for(int i = (int)(a) ; i > (int)(b) ; --i)
// NOT MY CODE, FROM https://pitsbuffersolution.com/
int N,K;
int b[2002][2002];
int s[4002][4002];
 
int get(int x , int y){
	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];
}
 
 // NOT MY CODE, FROM https://pitsbuffersolution.com/
 
int main(){
	cin >> N >> K;
	repp(i,0,N){
		int x,y;
		char c;
		cin >> x >> y >> c;
		if(c=='W') y += K;
		++b[x%(2*K)+1][y%(2*K)+1];
	}
	repp(i,1,4*K+1){
		repp(j,1,4*K+1){
			s[i][j] = s[i][j-1] + b[(i-1)%(2*K)+1][(j-1)%(2*K)+1];
		}
	}
	repp(j,1,4*K+1){
		repp(i,1,4*K+1){
			s[i][j] += s[i-1][j];
		}
	}
	int ans = 0;
	repp(i,1,2*K+1){
		repp(j,1,2*K+1){
			ans = max(ans , get(i,j) + get(i+K,j+K));
		}
	}
	cout << ans << endl;
	return 0;
}