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;

const int INF = 1e9;
const ll LLINF = 1e18;
const ll MOD = (ll)1e9 + 7;

int cal(int x1, int y1, int x2, int y2, vector<vector<int>>& imos)
{
  int ret =  imos[y2][x2] - imos[y2][x1] - imos[y1][x2] + imos[y1][x1];
  return ret;
}

int main()
{
  int N, K; scanf("%d %d", &N, &K);
  int x, y;
  char ch;
  vector<vector<int>> imos(K + 5, vector<int>(2 * K + 5, 0));
  for (int i = 0; i < N; ++i) {
    scanf("%d %d %c", &x, &y, &ch);
    if (ch == 'B') { x += K; }
    x %= 2 * K;
    y %= 2 * K;
    if (y >= K) {
      x = (x + K) % (2 * K);
      y %= K;
    }
    ++imos[y + 1][x + 1];
  }

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

  int ans = 0;
  for (int y = 1; y <= K; ++y) {
    for (int x = 1; x <= 2 * K; ++x) {
      int val = cal(max(0, x - K), max(0, y - K), x, y, imos) +
                cal(x, y, min(2 * K, x + K), K, imos) +
                cal(min(2 * K, x + K), max(0, y - K), 2 * K, y, imos) +
                cal(0, y, max(0, x - K), K, imos);
      ans = max(ans, val);
    }
  }
  printf("%d\n", ans);
  return 0;
}