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 black[2002][2002];
int Cumulative_sum[4002][4002]; //累積和
int k;

//累積和を二倍に取ることにより、二つの四角形で探索できるようになっている。
//ans = のところは、実質kの範囲の四角形二つに入っているBの数となっている。
//i=j=0の時にはget(i,j)でk～2k四角形、get(i+k,j+k)で0～k(2k～3k)の四角形を探索している。

int get(int x , int y){
    return Cumulative_sum[x+k-1][y+k-1] + Cumulative_sum[x-1][y-1] - Cumulative_sum[x-1][y+k-1] - Cumulative_sum[x+k-1][y-1];
}

int main() {

    int n;

    cin >> n >> k;

    int x,y;
    char c;

    for(int i=0; i<n; i++){
        cin >> x >> y >> c;
        if(c == 'W'){
            y += k;
        }
        black[x%(2*k)+1][y%(2*k)+1]++;
    }

    //累積和計算中
    for(int i=1; i<4*k+1; i++){
        for(int j=1; j<4*k+1; j++){
            Cumulative_sum[i][j] = Cumulative_sum[i][j-1] + black[(i-1)%(2*k)+1][(j-1)%(2*k)+1];
        }

    }

    //累積和計算中その2
    for(int j=1; j<4*k+1; j++){
        for(int i=1; i<4*k+1; i++){
            Cumulative_sum[i][j] += Cumulative_sum[i-1][j];
        }
    }

    int ans = 0;

    for(int i=1; i<2*k+1; i++){
        for(int j=1; j<2*k+1; j++){
            ans = max(ans, get(i,j) + get(i+k,j+k));
        }
    }

    cout << ans << endl;

    return 0;
}