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++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
#define fio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
using namespace std;
typedef long long int lli;
int K;
const int MM = 2020;
int matB[MM][MM], cumB[MM][MM];;
int cntB(int x1, int y1, int x2, int y2) {
    x2++; y2++;

    x1 = max(x1, 0);
    y1 = max(y1, 0);
    x2 = min(x2, 2 * K + 1);
    y2 = min(y2, 2 * K + 1);
    return cumB[x2][y2] + cumB[x1][y1] - cumB[x1][y2] - cumB[x2][y1];
}
int ctrB(int x, int y) {
    return cntB(x, y, x + K - 1, y + K - 1);
}
int main() {fio; // remove during scanf

    int N;
    cin >> N >> K;
    assert(K < 10);
    int totW = 0;
    for(int i = 0; i < N; i++) {
        int x, y;
        char c;
        cin >> x >> y >> c;
        x %= 2 * K;
        y %= 2 * K;
        if(c == 'B') {
            matB[x][y]++;
        } else {
            matB[x][y]--;
            totW++;
        }
    }
    for(int i = 0; i <= 2 * K; i++) {
        for(int j = 0; j <= 2 * K; j++) {
            cumB[i + 1][j + 1] = cumB[i + 1][j] + cumB[i][j + 1] - cumB[i][j] + matB[i][j];
        }
    }

    int ans = 0;
    for(int cx = 0; cx <= K; cx++) {
        for(int cy = 0; cy <= K; cy++) {
            int bcount = 0;
            bcount += ctrB(cx, cy);
            bcount += ctrB(cx + K, cy + K);
            bcount += ctrB(cx - K, cy + K);
            bcount += ctrB(cx + K, cy - K);
            bcount += ctrB(cx - K, cy - K);
            ans = max(ans, totW + bcount);
        }
    }
    // assert(ans > 1);
    cout << ans << endl;
    return 0;
}