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;

long long memo[2][1005][1005] = {0};
char c;
long long x, y, bw, n, k;

long long solve();

int main() {
  long long i, j, l;
  cin >> n >> k;
  for(i = 0; i < n; ++i) {
    cin >> x >> y >> c;
    bw = (c == 'B');
    ++memo[(x / k + y / k + bw) % 2][x % k][y % k];
  }
  for(l = 0; l < 2; ++l)
    for(i = 0; i < k; ++i)
      for(j = 0; j < k; ++j) {
        if(j != 0) memo[l][i][j] += memo[l][i][j - 1];
        if(i != 0) memo[l][i][j] += memo[l][i - 1][j];
        if(i != 0 && j != 0)
          memo[l][i][j] -= memo[l][i - 1][j - 1];
      }
  cout << solve() << endl;
  return 0;
}

long long solve() {
  long long ans = 1, now[2];
  for(int i = 0; i < k; ++i)
    for(int j = 0; j < k; ++j)
      for(int l = 0; l < 2; ++l) {
        now[l] = memo[l][i][j];
        if(i != k - 1 && j != k - 1) {
          now[l] += memo[l][k - 1][k - 1];
          now[l] -= memo[l][i][k - 1];
          now[l] -= memo[l][k - 1][j];
          now[l] += memo[l][i][j];
        }
        now[l] +=
            (memo[1 - l][k - 1][j] - memo[1 - l][i][j]);
        now[l] +=
            (memo[1 - l][i][k - 1] - memo[1 - l][i][j]);
        ans = max(ans, now[l]);
      }
  return ans;
}