Problem:
Score : 500 points

Problem Statement

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:

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.

Sample Input 1

4 3
0 1 W
1 2 W
5 3 B
5 4 B

Sample Output 1

4

He can satisfy all his desires by painting as shown in the example above.

Sample Input 2

2 1000
0 0 B
0 1 W

Sample Output 2

2

Sample Input 3

6 2
1 2 B
2 1 W
2 2 B
1 0 B
0 6 W
4 5 W

Sample Output 3

4

Solution in C (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>

inline void gc(int *x) {
    *x = 0;
    for (;;) {
        int c = getchar_unlocked();
        if ('0' <= c && c <= '9') {
            *x = *x * 10 + c - '0';
        } else {
            break;
        }
    }
}

#define KMAX (1000)
int s[KMAX + 1][2 * KMAX + 1];
int main(void) {
    char c;
    int n, k, x, y;
    scanf("%d %d\n", &n, &k);
    for (int i = 0; i < n; i++) {
        gc(&x);
        gc(&y);
        c = getchar_unlocked();
        getchar_unlocked();
        y %= 2 * k;
        if (c == 'W') {
            if (y >= k) {
                y -= k;
            } else {
                x += k;
            }
        } else {
            if (y >= k) {
                y -= k;
                x += k;
            }
        }
        x %= 2 * k;
        s[y + 1][x + 1]++;
    }
    for (int y = 0; y < k; y++) {
        for (int x = 0; x < 2 * k; x++) {
            s[y + 1][x + 1] += s[y][x + 1] + s[y + 1][x] - s[y][x];
        }
    }
    int max = 0;
    int v = s[0][0] - s[0][2 * k];
    for (int y = 0; y < k; y++) {
        int u = v - s[y][0] + s[y][2 * k];
        for (int x = 0; x < k; x++) {
            int t = u - s[0][x] - s[k][x] + s[0][x + k] + s[k][x + k] +
                    2 * s[y][x] - 2 * s[y][x + k];
            t = t < n / 2 ? n - t : t;
            max = t > max ? t : max;
        }
    }
    printf("%d", max);
    return 0;
}