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 ri register
#define int long long
#define E (n+1)
using namespace std; const int N=2010;
inline int read()
{
    int s=0, w=1; ri char ch=getchar();
    while(ch<'0'||ch>'9') { if(ch=='-') w=-1; ch=getchar(); }
    while(ch>='0'&&ch<='9') s=(s<<3)+(s<<1)+(ch^48), ch=getchar();
    return s*w;
}
int n,K,s[N][N]; char op[3];
inline int F(int x,int y,int p,int q)
{
    return s[p][q]-s[p][y]-s[x][q]+s[x][y];
}
signed main()
{
    n=read(), K=read();
    for(ri int i=1;i<=n;i++)
    {
        int x,y;
        scanf("%lld%lld%s",&x,&y,op);
        x%=(K*2), y%=(K*2);
        if(op[0]=='B') y+=K, y%=(K*2);
        s[x+1][y+1]++;
    }
    for(ri int i=1;i<=K+K;i++)
    for(ri int j=1;j<=K+K;j++)
    s[i][j]+=s[i-1][j]+s[i][j-1]-s[i-1][j-1];
    
    int res=0;
    for(ri int i=1;i<=K;i++)
    for(ri int j=1;j<=K;j++)
    {
        int qwq=0;
        qwq+=F(0,0,i,j)+F(i,j,i+K,j+K)+F(i+K,j+K,K*2,K*2)+F(i+K,0,K*2,j)+F(0,j+K,i,K*2);
        res=max(res,max(qwq,n-qwq));
    }
    printf("%lld\n",res);
    return 0;
}