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;

const int MAX_K = (int)3000 + 5;

int x[MAX_K][MAX_K];
int n, k;

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

int main() {
    scanf("%d %d", &n, &k);

    int a, b;
    char c;
    for (int i = 0; i < n; ++i) {
      scanf("%d %d %c", &a, &b, &c);
      if(c == 'W') b+=k;
      // a %= 2 * k, b %= 2 * k;
      x[a % (2 * k) + 1][b % (2 * k) + 1] += 1;
    }

    for (int i = 1; i <= 2 * k; ++i) for (int j = 1; j <= 2 * k; ++j) {
      x[i][j] = x[i][j] + (x[i - 1][j] + x[i][j - 1] - x[i - 1][j - 1]);
    }

    int ans = 0, dual, sum;

    for (int i = 1; i <= k; ++i) for (int j = 1; j <= k; ++j){
      sum = getsize(0, 0, i, j);
      sum += getsize(i, j, k + i, k + j);
      sum += getsize(k + i, k + j, 2 * k, 2 * k);
      sum += getsize(k + i, 0, 2 * k, j);
      sum += getsize(0 ,k + j, i, 2 * k);
      dual = n - sum;
      sum = max(dual, sum);
      ans = max(ans, sum);
    }

    printf("%d\n",ans);
    return 0;
}