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.

Example

Input

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


Output

2
2
10
-1
13
-1
import bisect

N,Q = (int(x) for x in input().split())

stop_arr = [tuple(map(int, input().split())) for _ in range(N)]
Q_arr = [int(input()) for _ in range(Q)]

stop_arr.sort(key=lambda x:x[2])

ans_arr = [-1]*Q
skip_arr = [-1]*(Q+1)

def update_skip(skip_arr, x):
    if skip_arr[x] == -1:
        return x
    x_dest = update_skip(skip_arr, skip_arr[x])
    skip_arr[x] = x_dest
    return x_dest

for s,t,x in stop_arr:
    q_start = bisect.bisect_left(Q_arr, s-x)
    q_stop = bisect.bisect_left(Q_arr, t-x)
    l = q_start
    while l < q_stop:
        if skip_arr[l] == -1:
            skip_arr[l] = q_stop
            ans_arr[l] = x
            l += 1
        else:
            l = update_skip(skip_arr, l)

print(*ans_arr, sep="\n")
