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

-----Sample Output-----
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.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = 'import heapq\n# sys.stdin.readline()\nimport sys\ninput = sys.stdin.readline\n\nN,Q = list(map(int, input().split()))\n\nS,T,X = [],[],[]\n\nEv = []\nfor _ in range(N):\n  s,t,x = list(map(int, input().split())) \n  Ev.append((s-x,True,x))\n  Ev.append((t-x,False,x))\n"""\n2 5 6\n5 -1\n4 2\n3 3\n2 4\n1 5\n0 -1\n"""\nEv.sort()\n#print(Ev)\n# C++のSetはminもO(logN)で取れるが、Pythonは無理。\n# setとheapqを併用すると計算量が実現できる。\nh = []\nclosed = set()\neidx = 0\nD = []\n\n# 受け取った方が早い・・・？\nfor i in range(Q):\n  D.append(int(input()))\n\nfor d in D:\n  # dはsortedでinputされる。\n  #d = int(input())\n  #Query.append(d)\n  while eidx < 2*N and Ev[eidx][0] <= d:\n    #pop\n    t,e,x = Ev[eidx]\n    eidx += 1\n    if e:\n      heapq.heappush(h, x)\n      closed.add(x)\n    else:\n      closed.remove(x)\n      \n  # 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\n  while h and h[0] not in closed:\n    heapq.heappop(h)\n    \n  # assert( len(h) == len(set) )\n  if h:\n    print((h[0]))\n  else:\n    print((-1))\n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13803