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>
#define rep(i, n) for (ll i = 0; i < (n); i++)
#define all(x) (x).begin(), (x).end()
using ll = long long;
using namespace std;
template <typename T>
using vec = std::vector<T>;

int main() {
    int N, K;
    cin >> N >> K;
    vec<vec<int>> bk(4 * K, vec<int>(4 * K));
    rep(i, N) {
        int x, y;
        char c;
        cin >> x >> y >> c;
        if (c == 'B') x += K;
        ++bk[x % (2 * K)][y % (2 * K)];
    }
    rep(i,4*K)rep(j,4*K){
        bk[i][j] = bk[i%(2*K)][j%(2*K)];
    }

    rep(i, 4 * K-1) rep(j, 4 * K) {
        bk[i + 1][j] += bk[i][j];
    }
    rep(i, 4 * K) rep(j, 4 * K-1) {
        bk[i][j+1] += bk[i][j];
    }


    int ans = 0;
    rep(i,2*K)rep(j,2*K){
        int tmp = 0;
        tmp+=bk[i+K-1][j+K-1];
        if(i>0) tmp-=bk[i-1][j+K-1];
        if(j>0) tmp-=bk[i+K-1][j-1];
        if(i>0&&j>0) tmp+=bk[i-1][j-1];

        tmp+=bk[i+2*K-1][j+2*K-1];
        tmp-=bk[i+K-1][j+2*K-1];
        tmp-=bk[i+2*K-1][j+K-1];
        tmp+=bk[i+K-1][j+K-1];

        ans = max(ans,tmp);
    }
    cout << ans << endl;
}