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>
#define pt(sth) cout << sth << "\n"
#define chmax(a, b) {if(a<b) a=b;}
#define chmin(a, b) {if(a>b) a=b;}
#define moC(a, s, b) (a)=((a)s(b)+MOD)%MOD
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
static const ll INF=1e18;
static const ll MOD=1e9+7;
static const ll MAX=101010;

/*
 for(i=0; i<N; i++)
   cin >> a[i];
*/

ll s[2222][2222];
ll f[2222][2222];

int main(void) {
  ll i, j, k;
  
  ll N, K;
  cin >> N >> K;
  ll x[MAX], y[MAX];
  
  
  for(i=0; i<N; i++) {
    ll x, y;
    char c;
    cin >> x >> y >> c;
    x%=K*2; y%=K*2;
    if(c=='W') (y+=K)%=K*2;
    x++; y++;
    f[x][y]++;
  }
  
  
  for(i=1; i<=K*2; i++) for(j=1; j<=K*2; j++)
    s[i][j]+=s[i][j-1]+f[i][j];
  for(j=1; j<=K*2; j++) for(i=1; i<=K*2; i++)
    s[i][j]+=s[i-1][j];
  
  ll ans=0;
  for(i=1; i<=K+1; i++) {
    for(j=1; j<=K+1; j++) {
      ll t=0;
      t+=s[i-1][j-1];
      t+=s[i-1][K*2]-s[i-1][j+K-1];
      t+=s[i+K-1][j+K-1]-s[i-1][j+K-1]-s[i+K-1][j-1]+s[i-1][j-1];
      t+=s[K*2][j-1]-s[i+K-1][j-1];
      t+=s[K*2][K*2]-s[K*2][j+K-1]-s[i+K-1][K*2]+s[i+K-1][j+K-1];
      chmax(ans, max(t, N-t));
    }
  }
  
  pt(ans);
  
}