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;
int dp[2001][2001];
int query(int x1, int y1, int x2, int y2) {
  return dp[y2][x2] - dp[y1][x2] - dp[y2][x1] + dp[y1][x1];
}
int main() {
  int N,K,answer=0,x,y;
  char c;
  cin >> N >> K;
  for (int i = 0; i < N; dp[y+1][x+1]++, i++) {
    cin >> x >> y >> c;
    if (c == 'B') {
      y += K;
    }
    x %= 2*K;
    y %= 2*K;
  }
  for (int i = 0; i <= 2*K; i++) {
    for (int j = 0; j < 2*K; j++) {
      dp[i][j+1] += dp[i][j];
    }
  }
  for (int j = 0; j <= 2*K; j++) {
    for (int i = 0; i < 2*K; i++) {
      dp[i+1][j] += dp[i][j];
    }
  }
  for (int i = 0; i <= K; i++) {
    for (int j = 0, k = 0; j <= K; j++, k = 0) {
      k += query(0, 0, i, j);
	  k += query(i + K, 0, 2 * K, j);
	  k += query(0, j + K, i, 2 * K);
	  k += query(i, j, i + K, j + K);
	  k += query(i + K, j + K, 2 * K, 2 * K);
      answer = max(answer,max(k,N-k));
    }
  }
  cout << answer;
}