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;
int N, K;
int A[2010][2010] = {};

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

    for(int i=1; i<=2*K; ++i) {
        for(int j=1; j<=2*K; ++j) {
            A[i][j] += A[i][j - 1];
        }
    }
    for(int i=1; i<=2*K; ++i) {
        for(int j=1; j<=2*K; ++j) {
            A[i][j] += A[i - 1][j];
        }
    }

    int res = 0;
    for(int i=1; i<=K; ++i) {
        for(int j=1; j<=K; ++j) {
            int tmp = 0;
            tmp += A[i - 1][j - 1];
            tmp += A[2 * K][j - 1] - A[i + K - 1][j - 1];
            tmp += A[i - 1][2 * K] - A[i - 1][j + K - 1];
            tmp += A[2 * K][2 * K] - A[i + K - 1][2 * K] - A[2 * K][j + K - 1] + A[i + K - 1][j + K - 1];
            tmp += A[i + K - 1][j + K - 1] - A[i - 1][j + K - 1] - A[i + K - 1][j - 1] + A[i - 1][j - 1];
            res = max(res, tmp);
            res = max(res, N - tmp);
        }
    }
    cout << res << endl;
}