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;
using ll=long long;
template<class T> using V = vector<T>;
template<class T, class U> using P = pair<T, U>;
using vll = V<ll>;
using vvll = V<vll>;

const ll MOD = (ll)1e9 + 7;
const ll HIGHINF = (ll)1e18;

int main() {
  ll n, k; cin >> n >> k;
  vvll checker(k+1, vll(k+1, 0));
  ll x, y; char c; bool is_b;
  for (ll i=0; i<n; i++) {
    cin >> x >> y >> c;
    x %= 2*k; y %= 2*k;
    is_b = c == 'B';
    if (x >= k) {x -= k; is_b=!is_b;}
    if (y >= k) {y -= k; is_b=!is_b;}
    if (is_b) {
      checker.at(0).at(0)++;
      checker.at(y+1).at(0)--;
      checker.at(0).at(x+1)--;
      checker.at(y+1).at(x+1)+=2;
    } else {
      checker.at(y+1).at(0)++;
      checker.at(0).at(x+1)++;
      checker.at(y+1).at(x+1)-=2;
    }
  }
  for (ll i=1; i<k; i++) {
    for (ll j=0; j<k+1; j++) {
      checker.at(j).at(i) += checker.at(j).at(i-1);
    }
  }
  for (ll i=1; i<k; i++) {
    for (ll j=0; j<k+1; j++) {
      checker.at(i).at(j) += checker.at(i-1).at(j);
    }
  }
  ll ans = 0;
  for (ll i=0; i<k; i++) for (ll j=0; j<k; j++) {
    ans = max(ans, max(checker.at(i).at(j), n-checker.at(i).at(j)));
  }
  cout << ans << endl;
  return 0;
}