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 <string>
#include <vector>
#include <climits>
#include <algorithm>
#include <set>
#include <unordered_map>

using namespace std;
int N, K;
int x, y;
char c;
int sum[2001][2001];
int max(int a, int b,int c) {
  return max(max(a, b), c);
}

int main() {
  cin >> N >> K;
  for (int i=0; i<N; i++) {
    cin >> x >> y >> c;
    if (c == 'W') {
      x += K;
    }
    x = x%(2*K);
    y = y%(2*K);
    sum[x+1][y+1]++;
  }

  for (int i=0; i<2*K+1; i++) {
    for (int j=1; j<2*K+1; j++) {
      sum[i][j] += sum[i][j-1];
    }
  }
  for (int j=0; j<2*K+1; j++) {
    for (int i=1; i<2*K+1; i++) {
      sum[i][j] += sum[i-1][j];
    }
  }

  int ans = 0;
  for (int i=0; i<=K; i++) {
    for (int j=0; j<=K; j++) {
      int a = sum[i][j];
      int b = sum[i+K][j+K] - sum[i][j+K] - sum[i+K][j] + sum[i][j];
      int c = sum[2*K][2*K] - sum[2*K][j+K] - sum[i+K][2*K] + sum[i+K][j+K];
      int d = sum[2*K][j] - sum[i+K][j];
      int e = sum[i][2*K] - sum[i][j+K];
      ans = max(ans, a+b+c+d+e, N-(a+b+c+d+e));
    }
  }
  cout << ans << endl;
}