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
import sys

sys.setrecursionlimit(10000)
INF = float('inf')

N, Q = list(map(int, next(sys.stdin).split()))
STX = [list(map(int, next(sys.stdin).split())) for _ in range(N)]
D = [int(next(sys.stdin)) for _ in range(Q)]

ans = [-1 for _ in range(Q)]
nexts = [-1 for _ in range(Q)]

for s, t, x in sorted(STX, key=lambda stx: (stx[2], stx[1])):
    # [lo, hi) に出発した人は x で止まる
    # x の小さい順にソート済みなのでここで確定
    lo = max(0, bisect.bisect_left(D, s - x))
    hi = min(len(D), bisect.bisect_left(D, t - x))

    p = lo
    while p < hi:
        if nexts[p] == -1:
            ans[p] = x
            nexts[p] = hi
            p += 1
        else:
            prev = p
            p = nexts[p]
            nexts[prev] = hi

print('\n'.join(map(str, ans)))
