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

using namespace std;

int v[2001][2001];
int n,k;
int getsize(int x1,int y1,int x2,int y2){
    return v[x2][y2] - v[x2][y1] - v[x1][y2] + v[x1][y1];
}
int main()
{
    scanf("%d%d",&n,&k);
    int a,b;
    char c;
    for(int i = 1;i <= n;i ++){
        scanf("%d%d %c",&a,&b,&c);
        a %= 2 * k,b %= 2 * k;
        if(c == 'W')
            b += k;
        v[a % (2 * k) + 1][b % (2 * k) + 1] ++;
    }
    for(int i = 1;i <= 2 * k;i ++)
        for(int j = 1;j <= 2 * k;j ++)
            v[i][j] = v[i][j] + (v[i - 1][j] + v[i][j - 1] - v[i - 1][j - 1]);
    int ans = 0,sum;
    for(int i = 1;i <= k;i++)
        for(int j = 1;j <= k;j++){
            sum = getsize(0,0,i,j);
            sum += getsize(i,j,k + i,k + j) + getsize(k + i,k + j,2 * k,2 * k) + getsize(k + i,0,2 * k,j) + getsize(0,k + j,i,2 * k);
            ans = max(ans,max(sum,n - sum));
    }
    printf("%d\n",ans);
}
