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
#include <iostream>
#include <utility>

using namespace std;

int ct[2000][2000];
int r[2000][2000];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, k;
    int x, y;
    char c;
    cin>>n>>k;
    int kk=k+k;
    for(int i=0; i<n; i++){
        cin>>x>>y>>c;
        if(c=='W'){
            x+=k;
        }
        ++ct[y%(kk)][x%(kk)];
        ++ct[(y+k)%(kk)][(x+k)%(kk)];
    }
    
    for(int i=0; i<kk; i++){
        r[i][0]=0;
        for(int j=0; j<k; j++) r[i][0]+=ct[i][j];
        for(int j=1; j<kk; j++){
            r[i][j]=r[i][(j+kk-1)%kk]-ct[i][(j+kk-1)%kk]+ct[i][(j+k-1)%kk];
        }
    }

    for(int i=0; i<kk; i++){
        ct[0][i]=0;
        for(int j=0; j<k; j++) ct[0][i]+=r[j][i];
        for(int j=1; j<kk; j++){
            ct[j][i]=ct[(j+kk-1)%kk][i]-r[(j+kk-1)%kk][i]+r[(j+k-1)%kk][i];
        }
    }

    int maxcount=0;
    for(int i=0; i<kk; i++){
        for(int j=0; j<kk; j++){
            maxcount = max(ct[i][j], maxcount);
        }
    }
    cout<<maxcount<<'\n';
    return 0;
}
