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: Wrong Answer)

import sys
input = sys.stdin.readline

def I(): return int(input())
def MI(): return map(int, input().split())
def LI(): return list(map(int, input().split()))

def main():
    import bisect

    """
    各工事場所について，
    Dがどのレンジなら引っかかるかを記録しておく．
    最初全部のDに関して，infまで行けるものとして，
    Xが小さい順に工事区間を見て，引っかかるならば区間更新．
    一度値を決定した場所は飛ばしながら更新するようにすればO(Q)
    nxtを入れながら進めば各ノードを2回しか見ないからO(Q)
    """

    
    N,Q=MI()
    xst2=[]#x,ごとに，Dがどの範囲なら引っかかるか(s~t)
    xi=[[0,0]for _ in range(N)]
    for i in range(N):
        s,t,x=MI()
        xst2.append([x,s-x,t-x])
        xi[i]=[x,i]
    xi.sort()
    
    
    D=[0]*Q
    for i in range(Q):
        D[i]=I()
        
    inf=10**10
    ans=[inf]*(Q+5)#すこし多めにとっておく
    nxt=[0]*(Q+5)#次に見るべき番号
    for i in range(Q+2):
        nxt[i]=i+1
        
    
    for i in range(N):
        ii=xi[i][1]
        xx=xst2[ii][0]
        
        #1index
        ss=bisect.bisect_left(D,xst2[ii][1])+1
        tt=bisect.bisect_left(D,xst2[ii][2])+1
        j=ss
        nxt[ss-1]=tt
        while j<tt:
            if ans[j]==inf:
                ans[j]=xx
            j=nxt[j]
            nxt[j]=tt+1#次ここにきたら，tt+1のところまで移動するようにしておく

    
        
    for i in range(Q):
        res=ans[i+1]
        if res==inf:
            print(-1)
        else:
            print(res)
            
        
    
        

main()