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;
typedef long double ld;
typedef pair<int, int> pii;
#define rep(i,n) for (int i=0; i<(n); i++)
#define reps(i,s,n) for (int i=(s); i<(n); i++)
#define rep1(i,n) for (int i=1; i<=(n); i++)
#define repr(i,s,n) for (int i=(s)-1; i>=(n); i--)
#define print(a) cout << (a) << endl
#define all(v) v.begin(), v.end()
#define rsort(v) sort(v.rbegin(), v.rend())


int main() {
  int N, K, M;
  cin >> N >> K;
  M = 2*K;
  vector<vector<int>> b(M+1,vector<int>(M+1,0));
  rep(i,N) {
    int x,y;
    string c;
    cin >> x >> y >> c;
    x %= M; y %= M;
    if (c == "W") {
      x += K;
      x %= M;
    }
    b[x+1][y+1]++;
  }
  rep1(i,M) {
    rep1(j,M) {
      b[i][j] += b[i][j-1];
    }
  }
  rep1(i,M) {
    rep1(j,M) {
      b[i][j] += b[i-1][j];
    }
  }
  int ans = 0;
  rep(i,K) {
    rep(j,K) {
      int tmp;
      tmp = b[i+K][M]+b[M][j+K]-b[i][M]-b[M][j]-2*b[i+K][j+K]+2*b[i][j+K]+2*b[i+K][j]-2*b[i][j];
      ans = max(ans,tmp);
      tmp = b[M][M]-tmp;
      ans = max(ans,tmp);
    }
  }
  print(ans);
  return 0;
}