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 heapq
N,Q = map(int, input().split())

S,T,X = [],[],[]

Ev = []
for _ in range(N):
  s,t,x = map(int, input().split())
  Ev.append((s-x,True,x))
  Ev.append((t-x,False,x))
"""
2 5 6
5 -1
4 2
3 3
2 4
1 5
0 -1
"""
Ev.sort()
#print(Ev)
# C++のSetはminもO(logN)で取れるが、Pythonは無理。
# setとheapqを併用すると計算量が実現できる。
h = []
closed = set()
eidx = 0
for i in range(Q):
  # dはsortedでinputされる。
  d = int(input())
  #Query.append(d)
  while eidx < 2*N and Ev[eidx][0] <= d:
    #pop
    t,e,x = Ev[eidx]
    eidx += 1
    if e:
      heapq.heappush(h, x)
      closed.add(x)
    else:
      closed.remove(x)
      
  # heqpqの最小値がsetに含まれている間、捨て続ければ、setの要素とheapqの要素が一致する
  while h and h[0] not in closed:
    heapq.heappop(h)
    
  # assert( len(h) == len(set) )
  if h:
    print(h[0])
  else:
    print(-1)
