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 N, K, dots[2001][2001], best;

int count(int x1, int y1, int x2, int y2) {
    return dots[x2][y2] - dots[x2][y1] - dots[x1][y2] + dots[x1][y1];
}

int main() {
    cin >> N >> K;

    for (int i = 1; i <= N; ++i) {
        char c;
        int x, y;
        cin >> x >> y >> c;

        if (c == 'W') {  // (x, y, 'W') == (x, y + K, 'B')
            y += K;
        }

        // (x, y) == (x, y + 2 * K) == (x + 2 * K, y)
        dots[x % (2 * K) + 1][y % (2 * K) + 1] += 1;
        // Now, 1 <= x, y <= 2 * K.
    }

    // accumulation
    for (int x = 1; x <= 2 * K; ++x) {
        for (int y = 1; y <= 2 * K; ++y) {
            dots[x][y] += dots[x - 1][y] + dots[x][y - 1] - dots[x - 1][y - 1];
        }
    }

    // For each painting.
    for (int x = 1; x <= K; ++x) {
        for (int y = 1; y <= K; ++y) {
            int sum = 0;
            sum += count(0, 0, x, y);
            sum += count(x, y, K + x, K + y);
            sum += count(K + x, K + y, 2 * K, 2 * K);
            sum += count(K + x, 0, 2 * K, y);
            sum += count(0, K + y, x, 2 * K);
            sum = max(N - sum, sum);
            best = max(best, sum);
        }
    }

    cout << best << endl;
}