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<iomanip>
//#include<cstdio>
#include<vector>
#include<map>
#include<queue>
#include<algorithm>
#include<cmath>
#include<cassert>
using namespace std;
typedef long long ll;
const int Nmax = 1e5;
int N, K, X[Nmax], Y[Nmax], C[Nmax];

int main(){
    cin >> N >> K;
    vector<vector<int> > v(2*K, vector<int>(2*K));
    for(int i=0; i<N; i++){
        ll x, y;
        char c;
        cin >> x >> y >> c;
        if(c=='B') y+=K;
        x = x%(2*K);
        y = y%(2*K);
        v[x][y] += 1;
        v[(x+K)%(2*K)][(y+K)%(2*K)] += 1;
    }
    vector<vector<int> > cum(2*K+1, vector<int>(2*K+1));
    for(int i=0; i<2*K; i++){
        for(int j=0; j<2*K; j++){
            cum[i+1][j+1] = v[i][j]+cum[i][j+1]+cum[i+1][j]-cum[i][j];
            // cout << i << " " << j << " " << v[i][j] << " " << cum[i][j] << endl;
            // cout << v[i][j] << " ";
        }
        // cout << endl;
    }

    int ans = 0;
    for(int i=0; i<=K; i++){
        for(int j=0; j<=K; j++){
            int s = cum[i+K][j+K]-cum[i][j+K]-cum[i+K][j]+cum[i][j];
            // cout << s << endl;
            ans = max(ans, s);
            ans = max(ans, N-s);
        }
    }
    cout << ans << endl;
    
    return 0;
}