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 <iostream>
#include <vector>
#include <cmath>

using namespace std;


int main() {
	int N, K;

	cin >> N >> K;
	vector<vector<int>> map(3 * K, vector<int>(3 * K, 0));
	vector<vector<int>> sum(3 * K, vector<int>(3 * K, 0));

	for (int i = 0; i < N; i++)
	{
		int x, y;
		char c;
		cin >> x >> y >> c;
		if (c == 'B') {
			y += K;
		}
		map[y % (2 * K)][x % (2 * K)]++;
		y += K; x += K;
		map[y % (2 * K)][x % (2 * K)]++;

	}

	//2K*2Kの盤面を3K*3Kに拡張
	for (int i = 0; i < 3 * K; i++) {
		for (int j = 0; j < 3 * K; j++)
		{
			map[i][j] = map[i % (2 * K)][j % (2 * K)];
		}
	}


	//累積和計算
	for (int i = 0; i < 3 * K; i++) {
		for (int j = 0; j < 3 * K; j++) {
			sum[i][j] = map[i][j];
			if (i - 1 >= 0) sum[i][j] += sum[i - 1][j];
			if (j - 1 >= 0) sum[i][j] += sum[i][j - 1];
			if (i - 1 >= 0 && j - 1 >= 0) sum[i][j] -= sum[i - 1][j - 1];
		}
	}

	int ans = 0;

	//白端点指定
	for (int i = 0; i < 2 * K; i++)
	{
		int tmp;
		for (int j = 0; j < 2 * K; j++)
		{
			tmp = sum[i + K - 1][j + K - 1];
			if (i - 1 >= 0) tmp -= sum[i - 1][j + K - 1];
			if (j - 1 >= 0) tmp -= sum[i + K - 1][j - 1];
			if (i - 1 >= 0 && j - 1 >= 0) tmp += sum[i - 1][j - 1];
			ans = max(tmp, ans);
		}
	}

	cout << ans << endl;

	return 0;
}