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;

int main() {
    int n, k;
    cin >> n >> k;
    vector<vector<int>> acc(2*k+1, vector<int>(2*k+1));
    for (int i = 0; i < n; i++) {
        int x, y;
        cin >> x >> y;
        char c;
        cin >> c;
        if (c == 'W') y += k;
        x %= 2*k, y %= 2*k;
        acc[x+1][y+1]++;
    }

    for (int i = 0; i + 1 <= 2*k; i++) {
        for (int j = 0; j + 1 <= 2*k; j++) {
            acc[i+1][j+1] += acc[i][j+1] + acc[i+1][j] - acc[i][j];
        }
    }

    int ans = 0;
    vector<pair<int, int>> v = { { 0, 0 }, { -k, -k }, { -k, k }, { k, k }, { k, -k } };
    for (int i = 0; i <= k; i++) {
        for (int j = 0; j <= k; j++) {
            int can = 0;
            for (auto& e : v) {
                int ax = i + e.first, ay = j + e.second;
                int bx = ax + k, by = ay + k;
                for (auto& f : { &ax, &ay, &bx, &by }) {
                    *f = max(0, *f);
                    *f = min(2*k, *f);
                }
                can += acc[bx][by] - acc[bx][ay] - acc[ax][by] + acc[ax][ay];
            }
            ans = max(ans, can);
            ans = max(ans, n - can);
        }
    }

    cout << ans << endl;
}