Problem:
Score : 500 points

Problem Statement

There is an infinitely long street that runs west to east, which we consider as a number line.

There are N roadworks scheduled on this street.
The i-th roadwork blocks the point at coordinate X_i from time S_i - 0.5 to time T_i - 0.5.

Q people are standing at coordinate 0. The i-th person will start the coordinate 0 at time D_i, continue to walk with speed 1 in the positive direction and stop walking when reaching a blocked point.

Find the distance each of the Q people will walk.

Constraints

All values in input are integers.

1 \leq N, Q \leq 2 \times 10^5

0 \leq S_i < T_i \leq 10^9

1 \leq X_i \leq 10^9

0 \leq D_1 < D_2 < ... < D_Q \leq 10^9

If i \neq j and X_i = X_j, the intervals [S_i, T_i) and [S_j, T_j) do not overlap.

Input

Input is given from Standard Input in the following format:

N Q
S_1 T_1 X_1
:
S_N T_N X_N
D_1
:
D_Q

Output

Print Q lines. The i-th line should contain the distance the i-th person will walk or -1 if that person walks forever.

Sample Input 1

4 6
1 3 2
7 13 10
18 20 13
3 4 2
0
1
2
3
5
8

Sample Output 1

2
2
10
-1
13
-1

The first person starts coordinate 0 at time 0 and stops walking at coordinate 2 when reaching a point blocked by the first roadwork at time 2.

The second person starts coordinate 0 at time 1 and reaches coordinate 2 at time 3. The first roadwork has ended, but the fourth roadwork has begun, so this person also stops walking at coordinate 2.

The fourth and sixth persons encounter no roadworks while walking, so they walk forever. The output for these cases is -1.

Solution in PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

def solve():
    from collections import defaultdict, namedtuple
    from operator import attrgetter
    import sys

    input = sys.stdin.readline
    Event = namedtuple('Event', 'time type x_coordinate')

    inf = 1 << 31

    n, q = map(int, input().split())

    e = []  # 通行止めと解除にクエリ分割
    for _ in range(n):
        s, t, x = map(int, input().split())
        # 区間[s, t)に通行止め
        e.append(Event(t - x, 0, x))  # 通行止め解除/0にすることで、解除と開始が同時刻の場合、先に解除される
        e.append(Event(s - x, 1, x))  # 通行止め開始/1にすることで、同上

    for _ in range(q):
        d = int(input())
        e.append(Event(d, 2, -1))  # 求値クエリ/2にすることで、通行止めの設定解除の反映後に到着する

    e.sort(key=attrgetter('time', 'type'))

    stop_cnt = defaultdict(int)
    stop_cnt[inf] = 1
    stop_set = {inf}
    curr = inf

    for evt in e:
        if evt.type == 0:
            stop_cnt[evt.x_coordinate] -= 1
            if not stop_cnt[evt.x_coordinate]:
                stop_set.discard(evt.x_coordinate)
                if curr == evt.x_coordinate:
                    curr = min(stop_set)
        elif evt.type == 1:
            stop_cnt[evt.x_coordinate] += 1
            if stop_cnt[evt.x_coordinate] == 1:
                stop_set.add(evt.x_coordinate)
            if evt.x_coordinate < curr:
                curr = evt.x_coordinate
        else:
            if curr == inf:
                print(-1)
            else:
                print(curr)


if __name__ == '__main__':
    solve()