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 Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

GC.disable
class MinHeap
  def initialize
    @arr = []
  end

  def insert(key)
    @arr.push(key)

    # sifting up
    i = @arr.size - 1
    until i == 0
      par = (i - 1) / 2
      if @arr[i] >= @arr[par]
        break
      else
        @arr[par], @arr[i] = @arr[i], @arr[par]
        i = par
      end
    end
  end

  alias push insert

  def peek
    if @arr.empty?
      fail 'empty'
    end

    @arr[0]
  end

  def size
    @arr.size
  end

  def empty?
    @arr.empty?
  end

  def _pop
    last = @arr.pop
    if @arr.empty?
      return
    end

    i = 0
    # sifting down
    while true
      child = i*2+1
      unless child < @arr.size # we've hit the bottom
        break
      end

      if child + 1 < @arr.size && @arr[child + 1] < @arr[child]
        child += 1
      end

      if @arr[child] >= last
        break
      else
        @arr[i] = @arr[child]
        i = child
      end

    end
    @arr[i] = last
  end

  def pop
    if @arr.empty?
      fail 'empty'
    end

    top = @arr[0]
    last = @arr.pop
    if @arr.empty?
      return top
    end

    i = 0
    # sifting down
    while true
      child = i*2+1
      unless child < @arr.size # we've hit the bottom
        break
      end

      if child + 1 < @arr.size && @arr[child + 1] < @arr[child]
        child += 1
      end

      if @arr[child] >= last
        break
      else
        @arr[i] = @arr[child]
        i = child
      end

    end
    @arr[i] = last

    return top
  end
end

n, q = gets.split.map(&:to_i)
sxt = []
n.times do
  s, t, x = gets.split.map(&:to_i)
  unless t - x <= 0
    sxt.push( ((s-x) << 60) | (x << 30) | (t-x) )
  end
end
sxt.sort!
ds = q.times.map { gets.to_i }

heap = MinHeap.new
ans = []

i = 0
ds.each do |d|
  while !heap.empty? && (heap.peek & ((1<<30) - 1)) <= d
    heap._pop
  end

  while i < sxt.size && ((tmp = sxt[i]) >> 60) <= d
    unless (tmp & ((1<<30) - 1)) <= d
      heap.push(tmp & ((1<<60) - 1))
    end
    i += 1
  end

  if heap.empty?
    ans << -1
  else
    ans << (heap.peek >> 30)
  end
end

puts ans