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 <iomanip>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <utility>
#include <tuple>
#include <algorithm>
#include <numeric>
#include <cstdio>
#include <cmath>
#define rep(i, n) for (int i = 0; i < (n); i++)
using namespace std;
using ll = long long;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

int grid[1010][1010][2], sum[1010][1010][2];

int main() {
  int N, K;
  cin >> N >> K;
  rep(i, N) {
    int x, y;
    char tmp;
    cin >> x >> y >> tmp;
    bool c = (tmp=='W' ? 0 : 1);
    grid[x%K][y%K][(x/K+y/K+c)%2]++;
  }
  rep(i, K) rep(j, K) rep(k, 2)
    sum[i+1][j+1][k] = sum[i+1][j][k]+sum[i][j+1][k]-sum[i][j][k]+grid[i][j][k];
  int ans = 0;
  for (int i=1; i<=K; i++) {
    for (int j=1; j<=K; j++) {
      rep(k, 2) {
        int res = sum[K][K][k]-sum[i][K][k]-sum[K][j][k]+sum[i][j][k]*2;
        res += sum[i][K][k^1]+sum[K][j][k^1]-sum[i][j][k^1]*2;
        chmax(ans, res);
      }
    }
  }
  cout << ans << endl;
  return 0;
}