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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
typedef tuple<ll, ll, ll> T;

int N, K;
vector<vector<int>> xy(2010, vector<int>(2010, 0));

int Get(int x1,int y1,int x2,int y2){ if(x1 > x2 || y1 > y2) return 0; return xy[x2][y2] - xy[x1 - 1][y2] - xy[x2][y1 - 1] + xy[x1 - 1][y1 - 1]; }

int main(void){
    
    scanf("%d %d", &N, &K);
    int KK = 2 * K;
    for(int i = 0; i < N; ++i){
        int a, b; char c; scanf("%d %d %c", &a, &b, &c);
        if(c == 'W') b += K;
        a %= KK, b %= KK; 
        ++xy[a + 1][b + 1];
    }
    
    for(int i = 1; i < KK + 2; ++i) for(int j = 1; j < KK + 2; ++j) xy[i][j + 1] += xy[i][j];
    for(int i = 1; i < KK + 2; ++i) for(int j = 1; j < KK + 2; ++j) xy[j + 1][i] += xy[j][i];
    
    int ans = 0;
    for(int x = 1; x <= K; ++x){
        for(int y = 1; y <= K; ++y){
            int temp = Get(x, y, x + K - 1, y + K - 1) + Get(x + K, y + K, KK, KK) + Get(1, 1, x - 1, y - 1) + Get(x + K, 1, KK, y - 1) + Get(1, y + K, x - 1, KK);
            ans = max(ans,max(temp,N-temp));
        }
    }
    printf("%d\n", ans);
    
    return 0;
}
