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:
//https://atcoder.jp/contests/abc086/submissions/3526242
#include "bits/stdc++.h"
#define rep(i,n) for(int i = 0; i < (n); ++i)
using namespace std;
typedef long long int ll;
typedef pair<int, int> P;
typedef pair<ll, int> LP;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

int n, k;
int sum[4005][4005], b[2005][2005];

int calc(int x, int y){
	return sum[y+k][x+k] - sum[y+k][x] - sum[y][x+k] + sum[y][x];
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	cin >> n >> k;
	rep(i,n){
		int x, y;
		char c;
		cin >> x >> y >> c;
		if(c == 'W') x += k;
		x %= 2*k;
		y %= 2*k;
		b[y][x]++;
	}
	rep(i,4*k)rep(j,4*k) sum[i][j] = b[i%(2*k)][j%(2*k)];
	rep(i,4*k){
		for(int j = 1; j < 4*k; ++j){
			sum[i][j] += sum[i][j-1];
		}
	}
	for(int i = 1; i < 4*k; ++i){
		rep(j,4*k){
			sum[i][j] += sum[i-1][j];
		}
	}
	int ans = 0;
	rep(i,2*k)rep(j,2*k){
		int score = calc(j, i) + calc(j + k, i + k);
		chmax(ans, score);
	}
	cout << ans << endl;
	return 0;
}