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;

#ifdef LOCAL
#define debug(var) cerr << #var << " = " << var << endl;
#else
#define debug(var)
#endif

void init() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
}

const int N = 1e5+34;
const int K = 1e3+23;

int n, k;
int x[N], y[N];
long long m[4*K][4*K];

int rect(int i, int j, int I, int J) {
	int x = i&&j?m[i-1][j-1]:0,
		y = i?m[i-1][J]:0,
		z = j?m[I][j-1]:0;
	return m[I][J]-z-y+x;
}

void solve() {
	scanf("%d%d", &n, &k);
	for (int i = 0; i < n; ++i) {
		int x, y; char c; scanf("%d%d %c", &x, &y, &c);
		if (c == 'W') x += k;
		x %= 2*k; y %= 2*k;
		++m[x][y];
		m[x+2*k][y] = m[x][y];
		m[x][y+2*k] = m[x][y];
		m[x+2*k][y+2*k] = m[x][y];
	}
	for (int i = 0; i < 4*k; ++i) {
		for (int j = 0; j < 4*k; ++j) {
			if (i) m[i][j] += m[i-1][j];
			if (j) m[i][j] += m[i][j-1];
			if (i && j) m[i][j] -= m[i-1][j-1];
		}
	}
	int ans = 0;
	for (int i = 0; i < 2*k; ++i) { // offset x
		for (int j = 0; j < 2*k; ++j) { // offset y
			ans = max(ans, rect(i, j, i+k-1, j+k-1) + rect(i+k, j+k, i+2*k-1, j+2*k-1));
		}
	}
	printf("%d", ans);
}

int main() {
	init();
	int t = 1; //scanf("%d", &t);
	while (t--) {
		solve();
	}
	return 0;
}
