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>
using namespace std;

#define K 1000
#define K2 K*2
#define K4 K*4

int cnt[K2][K2];
int sum[K4][K4];

int get( int x, int y, int k )
{
	return sum[x][y] + sum[x + k][y + k] - sum[x + k][y] - sum[x][y + k];
}

int main()
{
	int ans = 0;
	int n, k;
	int x, y;
	char c;

	cin >> n >> k;
	int k2 = k*2;
	for( int i = 0; i < n; i++ ) {
		cin >> x >> y >> c;
		if( c == 'W' ) {
			y += k;
		}
		x %= k2;
		y %= k2;
		cnt[x][y]++;
	}

	for(int i = 0; i < k2; i++ ) {
		for( int j = 0; j < k2; j++ ) {
			sum[i][j] = cnt[i][j];
			sum[i+k2][j] = cnt[i][j];
			sum[i][j+k2] = cnt[i][j];
			sum[i+k2][j+k2] = cnt[i][j];
		}
	}
	int k4 = k*4;
	for(int i = 0; i < k4; i++ ) {
		for( int j = 1; j < k4; j++ ) {
			sum[i][j] += sum[i][j - 1];
		}
	}
	for( int j = 0; j < k4; j++ ) {
		for( int i = 1; i < k4; i++ ) {
			sum[i][j] += sum[i - 1][j];
		}
	}

	for( int i = 0; i < k2; i++ ) {
		for( int j = 0; j < k2; j++ ) {
			int s = get( i, j, k ) + get( i + k, j + k, k );
			ans = max( ans, max( s, n - s ) );
		}
	}

	cout << ans << endl;

	return 0;
}
