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

// D - Checker

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
// const int INF = 2147483647;
// const ll INF = 9223372036854775807;
// const ll MOD = 1e9 + 7;

int N, K;
int h[3000][2000];
int cum[3001][2001];

int main() {
	cin >> N >> K;

	for (int i=0; i<N; i++) {
		int x, y;
		char c;
		cin >> x >> y >> c;

		// W to B
		if (c == 'W') x += K;

		// 0 < x < K*2, 0 < y < K の範囲に変換
		x = x % (K*2);
		y = y % (K*2);
		if (y >= K) {
			x = (x+K) % (K*2);
			y = y % K;
		}

		h[x][y]++;
		h[x+K][y+K]++;
		if (x-K >= 0) h[x-K][y+K]++;
		if (x+K*2 < K*3) h[x+K*2][y]++;
	}

	/*
	// **** debug ****
	for (int y=0; y<K*2; y++) {
		for (int x=0; x<K*3; x++) {
			cout << h[x][y] << " ";
		}
		cout << endl;
	}
	*/

	for (int y=0; y<K*2; y++) {
		for (int x=0; x<K*3; x++) {
			cum[x+1][y+1] = cum[x][y+1] + cum[x+1][y] - cum[x][y] + h[x][y];
		}
	}

	int ans = 0;
	for (int y=0; y<K; y++) {
		for (int x=0; x<K*2; x++) {
			int ok = cum[x+K][y+K] - cum[x][y+K] - cum[x+K][y] + cum[x][y];
			ans = max(ans, ok);
		}
	}

	cout << ans << endl;

	return 0;
}
