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 <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
#define mp make_pair

const int N = 2020;
int a[N][N];
int m, n;
int ans = 0;

int getSum(int x1, int y1, int x2, int y2) {
	return a[x2][y2] + a[x1][y1] - a[x2][y1] - a[x1][y2];
}

int solve(int x, int y) {
	if (x <= n)
		return getSum(x, y, x + n, n) + getSum(0, 0, x, y) + getSum(x + n, 0, 2 * n, y);
	else
		return getSum(0, y, x - n, n) + getSum(x - n, 0, x, y) + getSum(x, y, 2 * n, n);
}

int main()
{

	scanf("%d%d", &m, &n);
	while(m--) {
		int x, y;
		char c;
		scanf("%d%d %c", &x, &y, &c);
		if (c == 'W')
			x += n;
		x %= 2 * n;
		y %= 2 * n;
		if (y >= n) {
			x += n;
			y -= n;
		}
		x %= 2 * n;
		a[x + 1][y + 1]++;
	}

	for (int x = 1; x < N; x++)
		for (int y = 1; y < N; y++)
			a[x][y] += a[x - 1][y] + a[x][y - 1] - a[x - 1][y - 1];

	for (int x = 0; x < 2 * n; x++)
		for (int y = 0; y < n; y++) {
			ans = max(ans, solve(x, y));
		}
	printf("%d\n", ans);

	return 0;
}
