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 <string>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <math.h>
#include <cassert>
#define rep(i,n) for(int i = 0; i < n; ++i )
using namespace std;
using ll = long long;
using P = pair<int,int>;

const int K = 1000;
int v[2*K][2*K];
int s[2*K+1][2*K+1];
const int dx[] = {0,1,2,1,0,-1,-2,-1};
const int dy[] = {2,1,0,-1,-2,-1,0,1};
int main() {
  int n,k;
  cin >> n >> k;
  rep(i,n){
    int x,y; char c;
    cin >> x >> y >> c;
    if(c=='B') x+=k;
    x %= 2*k; y %= 2*k;
    v[x][y]++;
  }
  rep(i,2*k)rep(j,2*k) s[i+1][j+1] = s[i][j+1]+s[i+1][j]-s[i][j]+v[i][j];
  auto f = [&](int x1,int y1,int x2,int y2){
    if(x1>=2*k||y1>=2*k||x2<0||y2<0) return 0;
    x1=max(x1,0); y1=max(y1,0); x2=min(x2,2*k); y2=min(y2,2*k);
    return s[x2][y2] - s[x2][y1] - s[x1][y2] + s[x1][y1];
  };
  int ans = 0;
  rep(i,2*k)rep(j,k){
    int temp = f(i,j,i+k,j+k);
    rep(d,8) {
      int x = i+k*dx[d], y = j+k*dy[d];
      temp += f(x,y,x+k,y+k);
    }
    ans = max(ans,temp);
  }
  cout << ans << endl;
}