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>
using namespace std;

int cusum[2020][2020];

int main() {
    cin.tie(0); ios_base::sync_with_stdio(false);
    // in
    int N, K; cin >> N >> K;
    int x[101010], y[101010];
    char c[101010];
    for(int i=0; i < N; ++i) cin >> x[i] >> y[i] >> c[i];

    // calc
    // 基本方針：二次元累積和
    // しかしx, yは大きい。
    // 市松模様なので(0, 0)から2 * Kの位置で全探索すると
    // 他の場所の状態は一意にさだまる
    // つまりx, yをmod 2 * Kで場合分けすればよい？（白ならKずらす）
    for(int i=0; i < N; ++i) {
        if(c[i] == 'W') x[i] += K;
        int mx = x[i] % (K * 2);
        int my = y[i] % (K * 2);
        ++cusum[my  ][mx  ];
        ++cusum[(my+K) % (K * 2)][(mx+K) % (K * 2)];
    }

    for(int i = 0; i < K * 2; ++i) 
        for(int j = 0; j < K * 2; ++j) 
            cusum[i+1][j+1] = cusum[i+1][j+1] + cusum[i][j+1] + cusum[i+1][j] - cusum[i][j];
    
    // out
    int ans = 0;
    for(int i=0; i < K; ++i) {
        for(int j=0; j < K; ++j) {
            int unko = cusum[i+K][j+K] - cusum[i+K][j] - cusum[i][j+K] + cusum[i][j];
            ans = max(ans, unko);
            ans = max(ans, N-unko);
        }
    }

    cout << ans << endl;
    return 0;
}