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<vector>
#include<map>
#include<algorithm>
#include<string>
#include<iomanip>
#include<cstdio>

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)   FOR(i,0,n)
#define MOD 1000000007
#define MAXS(x,y) (x = max(x,y))
#define MINS(x,y) (x = min(x,y))

using namespace std;

typedef long long int ll;

const ll INF=(ll)1e18;


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];
}

int main(){
  cin >> N >> K;

  REP(i,N){
    ll xx,yy;
    char cc;
    cin >> xx >> yy >> cc;
    if(cc == 'W')yy += K;
    ++b[xx%(2*K) + 1][yy% (2*K) + 1];
  }

  FOR(i,1,4*K+1)FOR(j,1,4*K+1){
    s[i][j] += s[i][j-1] + b[(i-1)%(2*K)+1][(j-1)%(2*K)+1];;
  }
  FOR(i,1,4*K+1)FOR(j,1,4*K+1){
    s[j][i] += s[j-1][i];
  }

  int ans = 0;
	FOR(i,1,2*K+1){
		FOR(j,1,2*K+1){
			ans = max(ans , get(i,j) + get(i+K,j+K));
		}
	}

  cout << ans << endl;
}