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<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i = 0; i < (n); i++)
#define rep1(i, n) for(int i = 1; i <= (n); i++)
#define co(x) cout << (x) << "\n"
#define cosp(x) cout << (x) << " "
#define ce(x) cerr << (x) << "\n"
#define cesp(x) cerr << (x) << " "
#define pb push_back
#define mp make_pair
#define Would
#define you
#define please

int D[1001][1001];

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);


	int N, K;
	cin >> N >> K;

	int gyaku = 0;
	rep(i, N) {
		int x, y;
		char c;
		cin >> x >> y >> c;
		int k = x / K ^ y / K;
		x = x % K;
		y = y % K;
		if (c == 'W' ^ (k & 1)) {
			gyaku++;
			D[x][y]--;
		}
		else D[x][y]++;
	}

	for (int i = K - 1; i >= 0; i--) {
		for (int j = K - 1; j >= 0; j--) {
			D[i][j] = D[i][j + 1] + D[i + 1][j] - D[i + 1][j + 1] + D[i][j];
		}
	}
	int saidai = -1e9;
	rep(i, K) rep(j, K) {
		int kari = D[i][0] + D[0][j] - D[i][j] - D[i][j];
		saidai = max(saidai, kari);
		saidai = max(saidai, D[0][0] - kari);
	}
	co(gyaku + saidai);

	Would you please return 0;
}