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 Python (3.4.3) (judge verdict: Accepted)

# using main() makes code faster from the point of view of "access to variables in global name-space"
# def main():

""" convenient functions
# for i, a in enumerate(iterable)
# q, mod = divmod(a, b)
# divmod(x, y) returns the tuple (x//y, x%y)
# manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
"""

import sys
sys.setrecursionlimit(10**7)
from itertools import accumulate, combinations, permutations, product # https://docs.python.org/ja/3/library/itertools.html
# accumulate() returns iterator! to get list: list(accumulate())
from math import factorial, ceil, floor
def factorize(n):
    """return the factors of the Arg and count of each factor
    
    Args:
        n (long): number to be resolved into factors
    
    Returns:
        list of tuples: factorize(220) returns [(2, 2), (5, 1), (11, 1)]
    """
    fct = []  # prime factor
    b, e = 2, 0  # base, exponent
    while b * b <= n:
        while n % b == 0:
            n = n // b
            e = e + 1
        if e > 0:
            fct.append((b, e))
        b, e = b + 1, 0
    if n > 1:
        fct.append((n, 1))
    return fct
def combinations_count(n, r):
    """Return the number of selecting r pieces of items from n kinds of items.
    
    Args:
        n (long): number
        r (long): number
    
    Raises:
        Exception: not defined when n or r is negative
    
    Returns:
        long: number
    """
    # TODO: How should I do when n - r is negative?
    if n < 0 or r < 0:
        raise Exception('combinations_count(n, r) not defined when n or r is negative')
    if n - r < r: r = n - r
    if r < 0: return 0
    if r == 0: return 1
    if r == 1: return n
    numerator = [n - r + k + 1 for k in range(r)]
    denominator = [k + 1 for k in range(r)]
    for p in range(2,r+1):
        pivot = denominator[p - 1]
        if pivot > 1:
            offset = (n - r) % p
            for k in range(p-1,r,p):
                numerator[k - offset] /= pivot
                denominator[k] /= pivot
    result = 1
    for k in range(r):
        if numerator[k] > 1:
            result *= int(numerator[k])
    return result
def combinations_with_replacement_count(n, r):
    """Return the number of selecting r pieces of items from n kinds of items allowing individual elements to be repeated more than once.
    
    Args:
        n (long): number
        r (long): number
    
    Raises:
        Exception: not defined when n or r is negative
    
    Returns:
        long: number
    """
    if n < 0 or r < 0:
        raise Exception('combinations_with_replacement_count(n, r) not defined when n or r is negative')
    elif n == 0:
        return 1
    else:
        return combinations_count(n + r - 1, r)
from bisect import bisect_left, bisect_right
from collections import deque, Counter, defaultdict # https://docs.python.org/ja/3/library/collections.html#collections.deque
from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
from operator import itemgetter
# ex1: List.sort(key=itemgetter(1))
# ex2: sorted(tuples, key=itemgetter(1,2))
from functools import reduce
def chmin(x, y):
    """change minimum
    if x > y, x = y and return (x, True).
    convenient when solving problems of dp[i]
    
    Args:
        x (long): current minimum value
        y (long): potential minimum value
    
    Returns:
        (x, bool): (x, True) when updated, else (x, False)
    """
    if x > y:
        x = y
        return (x, True)
    else:
        return (x, False)
def chmax(x, y):
    """change maximum
    if x < y, x = y and return (x, True).
    convenient when solving problems of dp[i]
    
    Args:
        x (long): current maximum value
        y (long): potential maximum value
    
    Returns:
        (x, bool): (x, True) when updated, else (x, False)
    """
    if x < y:
        x = y
        return (x, True)
    else:
        return (x, False)

from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
def gcds(numbers):
    return reduce(gcd, numbers)
def lcm(x, y):
    return (x * y) // gcd(x, y)
def lcms(numbers):
    return reduce(lcm, numbers, 1)

# first create factorial_list
# fac_list = mod_factorial_list(n)
INF = 10 ** 18
MOD = 10 ** 9 + 7
modpow = lambda a, n, p = MOD: pow(a, n, p) # Recursive function in python is slow!
def modinv(a, p = MOD):
    # evaluate reciprocal using Fermat's little theorem:
    # a**(p-1) is identical to 1 (mod p) when a and p is coprime
    return modpow(a, p-2, p)
def modinv_list(n, p = MOD):
    if n <= 1:
        return [0,1][:n+1]
    else:
        inv_t = [0,1]
        for i in range(2, n+1):
            inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
        return inv_t
def modfactorial_list(n, p = MOD):
    if n == 0:
        return [1]
    else:
        l = [0] * (n+1)
        tmp = 1
        for i in range(1, n+1):
            tmp = tmp * i % p
            l[i] = tmp
        return l
def modcomb(n, k, fac_list = [], p = MOD):
    # fac_list = modfactorial_list(100)
    # print(modcomb(100, 5, modfactorial_list(100)))
    from math import factorial
    if n < 0 or k < 0 or n < k: return 0
    if n == 0 or k == 0: return 1
    if len(fac_list) <= n:
        a = factorial(n) % p
        b = factorial(k) % p
        c = factorial(n-k) % p
    else:
        a = fac_list[n]
        b = fac_list[k]
        c = fac_list[n-k]
    return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
def modadd(a, b, p = MOD):
    return (a + b) % MOD
def modsub(a, b, p = MOD):
    return (a - b) % p
def modmul(a, b, p = MOD):
    return ((a % p) * (b % p)) % p
def moddiv(a, b, p = MOD):
    return modmul(a, modpow(b, p-2, p))

""" initialize variables and set inputs
# initialize variables
    # to initialize list, use [0] * n
    # to initialize two dimentional array, use [[0] * N for _ in range(N)]
# set inputs
    # open(0).read() is a convenient method:
    # ex) n, m, *x = map(int, open(0).read().split())
    #     min(x[::2]) - max(x[1::2])
    # ex2) *x, = map(int, open(0).read().split())
    #     don't forget to add comma after *x if only one variable is used
# preprocessing
    # transpose = [x for x in zip(*data)]
    # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
    # flat = [flatten for inner in data for flatten in inner]
    # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [1, 2, 3, 4, 5, 6, 7, 8, 9]
# calculate and output
    # output pattern
    # ex1) print(*l) => when l = [2, 5, 6], printed 2 5 6
"""

# functions used
r = lambda: sys.stdin.readline().strip()
r_int = lambda: int(r())
R = lambda: list(map(int, r().split()))
Rfloat = lambda: list(map(float, r().split()))
Rtuple = lambda: tuple(map(int, r().split()))
Rmap = lambda: map(int, r().split())

""" how to treat input
# single int: int(r())
# single string: r()
# single float: float(r())
# line int: R()
# line string: r().split()
# line (str, int, int): [j if i == 0 else int(j) for i, j in enumerate(r().split())]
# lines int: [R() for _ in range(n)]
"""

# main
N, Q = R()
STX = [R() for _ in range(N)]
STX.sort(key=itemgetter(2))

D = [int(r()) for _ in range(Q)]
Stopped = [-1] * Q
ans = [-1] * Q

for s, t, x in STX:
    l = bisect_left(D, s-x)
    r = bisect_left(D,t-x)
    a = l
    while a < r:
        if Stopped[a] == -1:
            ans[a] = x
            Stopped[a] = r
            a += 1
        else:
            a = Stopped[a]

for i in ans:
    print(i)

"""memo: how to use defaultdict of list
# initialize
Dic = defaultdict(list)
# append / extend
Dic[x].append(y)
# for
for k, v in Dic.items():
"""

# if __name__ == '__main__':
#     main()