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 <bits/stdc++.h>
#define all(x) x.begin(), x.end()
#define sz(x) (int)x.size()
#define x first
#define y second
using namespace std;

const int N = 4007;

int n, k, x, y, pw[N][N], pb[N][N];
char c;

int sw(int i, int j) {
	return pw[i + k][j + k] - pw[i + k][j] - pw[i][j + k] + pw[i][j];
}
int sb(int i, int j) {
	return pb[i + k][j + k] - pb[i + k][j] - pb[i][j + k] + pb[i][j];
}

signed main() {
#ifdef LC
    assert(freopen("input.txt", "r", stdin));
#endif
    ios::sync_with_stdio(0), cin.tie(0);
	
	cin >> n >> k;
	while (n--) {
		cin >> x >> y >> c;
		x %= 2 * k; ++x;
		y %= 2 * k; ++y;
		for (int i = 0; i < 2; ++i) {
			for (int j = 0; j < 2; ++j) {
				if (c == 'W') {
					++pw[x + 2 * i * k][y + 2 * j * k];
				} else {
					++pb[x + 2 * i * k][y + 2 * j * k];
				}
			}
		}
	}
	for (int i = 1; i < N; ++i) {
		for (int j = 1; j < N; ++j) {
			pw[i][j] += pw[i - 1][j] + pw[i][j - 1] - pw[i - 1][j - 1];
			pb[i][j] += pb[i - 1][j] + pb[i][j - 1] - pb[i - 1][j - 1];
		}
	}

	int ans = 0;
	for (int i = 0; i < 2 * k; ++i) {
		for (int j = 0; j < 2 * k; ++j) {
			int cur = sb(i, j) + sb(i + k, j + k) + sw(i + k, j) + sw(i, j + k);
			ans = max(ans, cur);
		}
	}
	cout << ans << "\n";
    return 0;
}
