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 (int i = 0; i < n; i++)
#define rrep(i, n) for (int i = n - 1; i >= 0; i--)
using namespace std;
#define INF ((1<<30)-1)
#define LINF (1LL<<60)
#define EPS (1e-10)
typedef long long ll;
typedef pair<ll, ll> P;
const int MOD = 1000000007;
const int MOD2 = 998244353;

// ref : https://atcoder.jp/contests/abc086/submissions/7707107

int b[2005][2005];

int main(){
    int n, k;
    cin >> n >> k;
    
    rep(i, n){
        int x, y;
        char c;
        cin >> x >> y >> c;
        if (c == 'W') y += k;
        b[y%(2*k)+1][x%(2*k)+1]++;
    }
    rep(i, 2*k) rep(j, 2*k) b[i+1][j+1] += b[i+1][j] + b[i][j+1] - b[i][j];

    int ans = 0;

    rep(i, k+1) rep(j, k+1){
        int tmp = 0;
        tmp += b[i+k][j+k] - b[i+k][j] - b[i][j+k] + b[i][j];
        tmp += b[2*k][2*k] - b[i+k][2*k] - b[2*k][j+k] + b[i+k][j+k];
        tmp += b[i][j];
        tmp += b[i][2*k] - b[i][j+k];
        tmp += b[2*k][j] - b[i+k][j];
        ans = max({ans, tmp, n - tmp});
    }

    cout << ans << endl;

    return 0;
}