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:

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

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

  for(auto i = 1; i <= 2*K; ++i) for(auto j = 1; j <= 2*K; ++j) S[i][j] += S[i][j-1];
  for(auto j = 1; j <= 2*K; ++j) for(auto i = 1; i <= 2*K; ++i) S[i][j] += S[i-1][j];
  auto sum = [S](int y1, int x1, int y2, int x2) {
    return S[y2+1][x2+1] - S[y1][x2+1] - S[y2+1][x1] + S[y1][x1];
  };

  auto res = 0;
  for(auto i = 0; i < K; ++i) for(auto j = 0; j < K; ++j) {
    auto s = 0;
    s += sum(i, j, i+K-1, j+K-1);
    if(i && j) s += sum(0, 0, i-1, j-1);
    if(i!=2*K-1 && j) s += sum(i+K, 0, 2*K-1, j-1);
    if(i && j!=2*K-1) s += sum(0, j+K, i-1, 2*K-1);
    s += sum(i+K, j+K, 2*K-1, 2*K-1);
    res = max({res, s, sum(0,0,2*K-1,2*K-1)-s});
  }
  cout << res << endl;
}