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
// practice with Dukkha
#include <algorithm>
#include <iostream>

using namespace std;

const int K = 1000, K2 = K * 2, K4 = K2 * 2;

int bb[K4][K4];

int main() {
	int n, k; cin >> n >> k;
	int k2 = k * 2;
	int k4 = k2 * 2;
	while (n--) {
		int i, j; cin >> i >> j, i %= k2, j %= k2;
		string s; cin >> s;
		if (s == "W")
			i += (i >= k ? -k : k);
		bb[i][j]++;
		bb[k2 + i][j]++;
		bb[i][k2 + j]++;
		bb[k2 + i][k2 + j]++;
	}
	for (int i = 0; i < k4; i++)
		for (int j = 1; j < k4; j++)
			bb[i][j] += bb[i][j - 1];
	for (int i = 1; i < k4; i++)
		for (int j = 0; j < k4; j++)
			bb[i][j] += bb[i - 1][j];
	int ans = 0;
	for (int i_ = 0; i_ < k2; i_++)
		for (int j_ = 0; j_ < k2; j_++) {
			int i, j, cnt = 0;
			i = i_, j = j_;
			cnt += bb[i + k][j + k] - bb[i][j + k] - bb[i + k][j] + bb[i][j];
			i = i_ + k, j = j_ + k;
			cnt += bb[i + k][j + k] - bb[i][j + k] - bb[i + k][j] + bb[i][j];
			ans = max(ans, cnt);
		}
	cout << ans << '\n';
	return 0;
}
