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 <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iomanip>

using namespace std;
struct color
{
    int x,y,no;
}s[100000];
int n,k,ans,sum;
int v[2000];
int main()
{
    char ch;
    scanf("%d%d",&n,&k);
    for(int i = 0;i < n;i ++)
    {
        scanf("%d %d %c",&s[i].x,&s[i].y,&ch);
        s[i].x %= k * 2;///通过移位（移动2k）都可以在这个2k*2k的图里进行操作 不改变原来的意义
        s[i].y %= k * 2;
        s[i].no = (ch == 'W' ? 1 : 0);///颜色转数字
    }
    for(int i = 0;i < k;i ++)///x坐标衡量
    {
        sum = 0;
        memset(v,0,sizeof(v));///初始化
        for(int j = 0;j < n;j ++)
        {
            int p = s[j].x >= i && s[j].x < i + k;///x坐标在i 到 i + k为一个区域
            int q = s[j].y >= 0 && s[j].y < k;///y坐标先不做限制 起点可以是0 ~ k - 1
            if(p ^ q ^ s[j].no)sum ++,v[s[j].y] ++;///先选定一种情况 满足的话 sum ++
            else v[s[j].y] --;///不满足的话
        }
        ans = max(ans,max(sum,n - sum));
        for(int j = 0;j < k;j ++)
        {
            sum -= v[j] + v[j + k];
            ans = max(ans,max(sum,n - sum));///更新最大值，sum 和 n - sum 是黑白互换的不同结果（原来黑色的地方可以变白色，那么白色的变黑色）
        }
    }
    printf("%d",ans);
}