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
#include <iostream>

using namespace std;

int order[2001][2001];

int sum_square(int x1, int y1, int x2, int y2) {
	return order[x2][y2] - order[x2][y1] - order[x1][y2] + order[x1][y1];
}

int main(int argc, char *argv[]) {
	int N, K;
	cin >> N >> K;

	for (int i = 0; i < N; i++) {
		int x, y;
		char c;
		cin >> x >> y;
		cin >> c;
		if (c == 'W') {
			x += K;
		}
		order[x % (2 * K) + 1][y % (2 * K) + 1] += 1;
	}

	for (int x = 1; x <= 2 * K; x++) {
		for (int y = 1; y <= 2 * K; y++) {
			order[x][y] += order[x - 1][y] + order[x][y - 1] - order[x - 1][y - 1];
		}
	}

	int ans = 0;
	for (int x = 1; x <= K; x++) {
		for (int y = 1; y <= K; y++) {
			int cnt = 0;
			cnt += sum_square(0, 0, x, y);
			cnt += sum_square(x, y, x + K, y + K);
			cnt += sum_square(x + K, y + K, 2 * K, 2 * K);
			cnt += sum_square(x + K, 0, 2 * K, y);
			cnt += sum_square(0, y + K, x, 2 * K);
			cnt = max(N - cnt, cnt);
			ans = max(ans, cnt);
		}
	}

	cout << ans << endl;
	return 0;
}