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;
typedef long long ll;
#define rep(i, n) for(ll i = 0; i < (ll)(n); i++)
#define rep1(i, n) for(ll i = 1; i < (ll)(n); i++)
const ll mo=1000000007;
const ll INF=1<<30;
ll ans=0;
int num[2010][2010]={};

struct edge {
    int x,y;
    char c;
};
vector<edge> ta;
int main()
{ 
ll n,k;
cin >> n>>k; 
for (int i = 0; i < n; i++) {
    int x, y;
    char D;
    cin >> x >> y >> D;
    x=x%(2*k);
    y=y%(2*k);
    if(y>=k){
        x=(x+k)%(2*k);
        y=y-k;
    }
    if(D=='W'){
        x=(x+k)%(2*k);
    }
    ta.push_back({ x,y,D });
}



rep(i,n){
    num[ta[i].y][ta[i].x]++;
    num[ta[i].y+k][(ta[i].x+k)%(2*k)]++;
}
rep(i,2*k)rep1(j,2*k){
    num[i][j]+=num[i][j-1];
}
rep1(i,2*k)rep(j,2*k){
    num[i][j]+=num[i-1][j];
}
ll temp;

rep(i,k)rep(j,k){
    ll p=0,q=0,r=0;
    if(i!=0&&j!=0)r=num[i-1][j-1];
    if(i!=0)p=num[i-1][j+k-1];
    if(j!=0)q=num[i+k-1][j-1];
    temp=num[i+k-1][j+k-1]+r-p-q;
    ans=max(ans,max(temp,n-temp));
}

/* for (int i = 0; i < 2*k; ++i){
for (int j = 0; j < 2*k; ++j) {
    cout << num[i][j] << " ";
}cout  << endl;} */
cout << ans << endl;
return 0;
} 