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>
using namespace std;

const int MAXN = 1010;
int G[MAXN*2][MAXN*2];

int calc(int x1, int y1, int x2, int y2)
{
    return G[x2][y2] - G[x1][y2] - G[x2][y1] + G[x1][y1];
}

int main()
{
    int n,k,x,y;
    char ch;
    scanf("%d %d",&n,&k);
    for(int i = 0; i < n; ++i)
    {
        scanf("%d %d %c",&x,&y,&ch);
        x = x%(2*k);
        y = y%(2*k);
        if(ch == 'W')
            y += k;
        ++G[x%(2*k)+1][y%(2*k)+1];
    }
    for(int i = 1; i <= 2*k; ++i)
        for(int j = 1; j <= 2*k; ++j)
            G[i][j] = G[i][j] + G[i-1][j] + G[i][j-1] - G[i-1][j-1];

    int res = 0, sum;
    for(int i = 1; i <= k; ++i)
    {
        for(int j = 1; j <= k; ++j)
        {
            sum = calc(0,0,i,j);
            sum += calc(i,j,i+k,j+k);
            sum += calc(i+k,j+k,2*k,2*k);
            sum += calc(i+k,0,2*k,j);
            sum += calc(0,j+k,i,2*k);
            sum = max(sum,n-sum);
            res = max(res,sum);
        }
    }
    printf("%d\n",res);
    return 0;
}