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;

#define rep(i, a, b) for (ll i = (a); i < (b); i++)
typedef uint64_t ull;
typedef int64_t ll;
typedef std::pair<ll, ll> PLL;

ll N,K;

// v[y][x]: 座標y,xに黒い点がいくつあるか
vector<vector<ll>> v(4100, vector<ll>(4100,0));

ll sum(ll x, ll y, ll w, ll h){
    return v[y+h][x+w]
          -v[y+h][x]
          -v[y][x+w]
          +v[y][x];
}

signed main() {
    cin>>N>>K;
    rep(i,0,N){
        ll x,y;
        char c;
        cin>>x>>y>>c;
        if (c=='W'){
            c='B';
            x += K;
        }
        x %= 2*K;
        y %= 2*K;
        v[y][x]++;
    }

    rep(y,0,2*K){
        rep(x,0,2*K){
            v[y+2*K][x+2*K] = v[y][x];
            v[y+2*K][x    ] = v[y][x];
            v[y    ][x+2*K] = v[y][x];
            v[y    ][x    ] = v[y][x];
        }
    }

    // v[y][x]を2次元累積和に変形
    rep(y,1,4*K)
        rep(x,0,4*K)
            v[y][x] += v[y-1][x];
    rep(x,1,4*K)
        rep(y,0,4*K)
            v[y][x] += v[y][x-1];

    ll ans = 0;
    rep(y,0,2*K)
        rep(x,0,2*K) {
            ll t = 0;
            t += sum(x,y,K,K);
            t += sum(x+K,y+K,K,K);
            ans = max(ans,t);
        }
    cout<<ans<<endl;
    return 0;
}