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
def d_checker(N, K, Q):
    # マス(x,y)が白⇔マス(x,y+K)が黒なので、希望の色がすべて黒になるようにする
    # 市松模様の周期は2Kなので、x,y座標をそれぞれ2Kで剰余を取った座標と同一視できる
    mat = [[0]*(2*K+2) for _ in range(2*K+2)]
    total = [[0]*(4*K+2) for _ in range(4*K+2)]
    for x, y, c in Q:
        if c == 'W':
            y += K
        mat[(x % (2*K)) + 1][(y % (2*K)) + 1] += 1

    # 事前に累積和を取る
    for x in range(1, 4*K+1):
        for y in range(1, 4*K+1):
            total[x][y] = total[x][y-1] + \
                mat[((x-1) % (2*K)) + 1][((y-1) % (2*K)) + 1]
    for y in range(1, 4*K+1):
        for x in range(1, 4*K+1):
            total[x][y] += total[x-1][y]

    def get(x, y):
        # 市松模様の橋の黒の座標を決めたときに実現できる希望の数
        return total[x+K-1][y+K-1] + total[x-1][y-1] - total[x-1][y+K-1] - total[x+K-1][y-1]

    # 正方形の隅の座標を各々決めたときに実現できる希望の数を求め、最大値を得る
    ans = 0
    for x in range(1, 2*K+1):
        for y in range(1, 2*K+1):
            ans = max(ans, get(x, y)+get(x+K, y+K))
    return ans

N, K = [int(i) for i in input().split()]
Q = [(int(x), int(y), c)
    for x, y, c in [input().split() for i in range(N)]]
print(d_checker(N, K, Q))
