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;

#define ll long long
#define pq priority_queue
#define all(v) (v).begin(), (v).end()
#define rep(i, n, m) for (int (i) = (n); (i) < (m); (i)++)
#define debug(x) cerr << #x << ": " << x << endl;
#define line() cerr << "---------------" << endl;

int N, K;
int sumB[1111][1111];
int sumW[1111][1111];

int main() {
   cin >> N >> K;
   rep (i, 0, N) {
      int x, y;
      char c;
      cin >> x >> y >> c;

      int tmp = x/K + y/K;
      x %= K, y%= K;
      if (tmp&1) {
         if (c=='B') c='W';
         else c = 'B';
      }
      if (c=='B') sumB[y+1][x+1]++;
      else sumW[y+1][x+1]++;
   }

   rep (i, 0, K) {
      rep (j, 0, K) {
         sumB[i+1][j+1]+=sumB[i][j+1]+sumB[i+1][j]-sumB[i][j];
         sumW[i+1][j+1]+=sumW[i][j+1]+sumW[i+1][j]-sumW[i][j];
      }
   }

   int ans = 0;
   rep (i, 0, K+1) {
      rep (j, 0, K+1) {
         int tmp = 0;
         tmp += sumB[i][j];
         tmp += sumB[K][K]-sumB[i][K]-sumB[K][j]+sumB[i][j];
         tmp += sumW[K][j]-sumW[i][j];
         tmp += sumW[i][K]-sumW[i][j];
         ans = max(ans, tmp);
         ans = max(ans, N-tmp);
      }
   }
   cout << ans << endl;

   return 0;
}