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>

using namespace std;

typedef long long ll;

int sum[2010][2010] = {};
int dot[2010][2010] = {};

int findDot(int x1, int y1, int x2, int y2)
{
    if (x2 < x1 || y2 < y1) return 0;
    return sum[x2][y2] - sum[x2][y1-1] - sum[x1-1][y2] + sum[x1-1][y1-1];
}

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

    int x, y;
    char c;
    for (int i = 0; i < N; ++i)
    {
        cin >> x >> y >> c;

        if (c == 'W') y += K;

        dot[x % (2*K) + 1][y % (2*K) + 1] += 1;
    }

    for (int i = 0; i < 2*K + 2; ++i)
    {
        for (int j = 0; j < 2*K + 2; ++j)
        {
            sum[i+1][j+1] = sum[i+1][j] + sum[i][j+1] - sum[i][j] + dot[i+1][j+1];
        }
    }

    int ans = 0;
    int tmp = 0;
    for (int x = 1; x <= K; ++x)
    {
        for (int y = 1; y <= K; ++y)
        {
            tmp = findDot(x, y, x+K-1, y+K-1)
                + findDot(x+K, y+K, 2*K, 2*K)
                + findDot(1, 1, x-1, y-1)
                + findDot(1, y+K, x-1, 2*K)
                + findDot(x+K, 1, 2*K, y-1);
            ans = max(ans, max(tmp, N-tmp));
        }
    }
    cout << ans << endl;

    return 0;
}