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;

int n, k;
int imos[1000*4][1000*4];

int main() {

    // input
    cin >> n >> k;
    while(n--){
        int x, y;
        char c;
        cin >> x >> y >> c;
        if(c == 'W') x += k;
        x %= 2*k;
        y %= 2*k;

        // +1, -1 for 2D-imos
        for(int i=0;i<4;i++){
            for(int j=0;j<4;j++){
                int xx = x%k+i*k+1;
                int yy = y%k+j*k+1;
                if((xx+yy)%(2*k) != (x+1+y+1)%(2*k)) continue;
                if(xx+k < 4*k && yy+k < 4*k){
                    imos[xx][yy]++;
                    imos[xx+k][yy]--;
                    imos[xx][yy+k]--;
                    imos[xx+k][yy+k]++;
                }
            }
        }
    }

    // 2D-imos
    int ans = 0;
    for(int i=0;i<4*k;i++){
        for(int j=1;j<4*k;j++){
            imos[i][j] += imos[i][j-1];
        }
    }
    for(int i=1;i<4*k;i++){
        for(int j=0;j<4*k;j++){
            imos[i][j] += imos[i-1][j];
            ans = max(ans, imos[i][j]);
        }
    }

    // output
    cout << ans << endl;

    return 0;
}