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 <cstdio>
using namespace std;

int max(int a, int b) { return a < b ? b : a; }

int board[2010][2010];

int main() {
    int N, K; scanf("%d%d", &N, &K);

    int sum = 0, white = 0;
    for(int i=0; i<N; i++) {
        int x, y; char c; scanf(" %d%d %c", &x, &y, &c);
        if(c=='W')y+=K;
        x %= 2*K, y %= 2*K;
        board[x+1][y+1]++;
    }

// ここから２次元累積和
    for(int i=1; i<=2*K; i++) {
        for(int j=1; j<=2*K; j++) {
            board[i][j] += board[i][j-1];
        }
    }

    for(int j=1; j<=2*K; j++) {
        for(int i=1; i<=2*K; i++) {
            board[i][j] += board[i-1][j];
        }
    }

// ここまで二次元累積和

    int ma = 0;
    for(int i=0; i<=K; i++) {
        for(int j=0; j<=K; j++) {
            int a = board[i][j];
            int b = board[i+K][j+K] - board[i+K][j] - board[i][j+K] + board[i][j];
            int c = board[2*K][j] - board[i+K][j];
            int d = board[i][2*K] - board[i][j+K];
            int e = board[2*K][2*K] - board[2*K][j+K] - board[i+K][2*K] + board[i+K][j+K];
            int tmp = a + b + c + d + e;
            ma = max(ma,max(tmp,N-tmp));
        }
    }
    printf("%d\n", ma);
    return 0;
}