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: Wrong Answer)

#define _CRT_SECURE_NO_WARNINGS

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <math.h>
#include <queue>
#include <stack>
#include <list>
#include <functional>
#include <numeric>
#include <iostream>

using namespace std;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < (n); ++i)

int W[2345][2345] = {}, B[2345][2345] = {};

int sum(int x1, int y1, int x2, int y2, bool b) {
	int a = 0;
	if (b) {
		if (x2 >= 0 && y2 >= 0) a += B[x2][y2];
		if (x1 - 1 >= 0 && y1 - 1 >= 0) a += B[x1 - 1][y1 - 1];
		if (x2 >= 0 && y1 - 1 >= 0) a -= B[x2][y1 - 1];
		if (x1 - 1 >= 0 && y2 >= 0) a -= B[x1 - 1][y2];
		return a;
	}
	else {
		if (x2 >= 0 && y2 >= 0) a += W[x2][y2];
		if (x1 - 1 >= 0 && y1 - 1 >= 0) a += W[x1 - 1][y1 - 1];
		if (x2 >= 0 && y1 - 1 >= 0) a -= W[x2][y1 - 1];
		if (x1 - 1 >= 0 && y2 >= 0) a -= W[x1 - 1][y2];
		return a;
	}
}

int main() {

	int N, K;

	scanf("%d%d", &N, &K);

	rep(i, N) {
		int x, y;
		char c;
		scanf("%d%d %c", &x, &y, &c);
		if (c == 'W') {
			W[x % (K * 2)][y % (K * 2)]++;
		}
		else {
			B[x % (K * 2)][y % (K * 2)]++;
		}
	}

	rep(i, (K * 2)) {
		rep(j, (K * 2)) {
			W[i][j + 1] += W[i][j];
			B[i][j + 1] += B[i][j];
		}
	}
	rep(i, (K * 2)) {
		rep(j, (K * 2)) {
			W[i + 1][j] += W[i][j];
			B[i + 1][j] += B[i][j];
		}
	}

	int ans = 0;
	rep(i, K) {
		rep(j, K) {
			int c = 0;
			c += sum(0, 0, i - 1, j - 1, true);
			c += sum(i, 0, i - 1 + K, j - 1, false);
			c += sum(i - 1 + K + 1, 0, K * 2 - 1, j - 1, true);
			c += sum(0, j, i - 1, j - 1 + K, false);
			c += sum(i, j, i - 1 + K, j - 1 + K, true);
			c += sum(i - 1 + K + 1, j, K * 2 - 1, j - 1 + K, false);
			c += sum(0, j - 1 + K + 0, i - 1, K * 2 - 1, true);
			c += sum(i, j - 1 + K + 1, i - 1 + K, K * 2 - 1, false);
			c += sum(i - 1 + K + 1, j - 1 + K + 1, K * 2 - 1, K * 2 - 1, true);
			ans = max(ans, c);
		}
	}

	printf("%d\n", ans);

	return 0;
}