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;
#define rep(i, N) for (int i = 0; i < (int)N; i++)
#define all(a) (a).begin(), (a).end()

int N, K;

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

  int T = 2*K;
  vector<vector<int>> C(T, vector<int>(T, 0));

  rep(i,N) {
    int x, y;
    char c;
    cin >> x >> y >> c;
    if(c == 'W') x += K;
    C[y%T][x%T]++;
  }

  vector<vector<int>> SC(T+1, vector<int>(T+1, 0));
  rep(i,T) rep(j,T) {
    SC[i+1][j+1] = C[i][j] + SC[i+1][j] + SC[i][j+1] - SC[i][j];
  }

  // [x0, x1) x [y0, y1)
  auto rangeSum = [&](int y0, int y1, int x0, int x1) {
    return SC[y1][x1] - SC[y0][x1] - SC[y1][x0] + SC[y0][x0];
  };

  int ans = 0;
  rep(i,T) rep(j,T) {
    int x1 = (j+K) % T, x2 = j;
    if(x1 > x2)swap(x1, x2);
    int y1 = (i+K) % T, y2 = i;
    if(y1 > y2)swap(y1, y2);

    int b = rangeSum(0, y1, 0, x1)
      + rangeSum(0, y1, x2, T)
      + rangeSum(y1, y2, x1, x2)
      + rangeSum(y2, T, 0, x1)
      + rangeSum(y2, T, x2, T);

    ans = max(ans, max(b, N - b));
  }

  cout << ans << endl;
  return 0;
}