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 LuaJIT (2.0.4) (judge verdict: Accepted)

local mce, mfl, msq, mmi, mma = math.ceil, math.floor, math.sqrt, math.min, math.max
local ior = io.input()
local n, q = ior:read("*n", "*n")
local s, t, x, d = {}, {}, {}, {}
for i = 1, n do
  s[i], t[i], x[i] = ior:read("*n", "*n", "*n")
end

local timelist = {}
local omitinfo = {}
for i = 1, q do
  local tmp = ior:read("*n")
  d[i] = tmp
  table.insert(timelist, tmp)
  local tmptbl = {}
  table.insert(omitinfo, tmptbl)
end
-- table.sort(timelist)

local function lower_bound(ary, x)
  if(x <= ary[1]) then return 1 end
  local num = #ary
  if(ary[num] < x) then return num + 1 end
  local min, max = 1, num
  while(1 < max - min) do
    local mid = mfl((min + max) / 2)
    if(ary[mid] < x) then
      min = mid
    else
      max = mid
    end
  end
  return max
end

local function upper_bound(ary, x)
  if(x < ary[1]) then return 1 end
  local num = #ary
  if(ary[num] <= x) then return num + 1 end
  local min, max = 1, num
  while(1 < max - min) do
    local mid = mfl((min + max) / 2)
    if(ary[mid] <= x) then
      min = mid
    else
      max = mid
    end
  end
  return max
end

for i = 1, n do
  -- [ngfrom, ngto]
  local ngfrom, ngto = s[i] - x[i], t[i] - x[i] - 1
  local real_ngfrom_pos_idx = lower_bound(timelist, ngfrom)
  local real_ngto_pos_idx = upper_bound(timelist, ngto) - 1
  -- print(i, ngfrom, ngto, real_ngfrom_pos_idx, real_ngto_pos_idx)
  if(real_ngfrom_pos_idx <= q and 1 <= real_ngto_pos_idx and real_ngfrom_pos_idx <= real_ngto_pos_idx) then
    table.insert(omitinfo[real_ngfrom_pos_idx], x[i])
    table.insert(omitinfo[real_ngto_pos_idx], -x[i])
  end
end
-- for i = 1, q do
--   print(i, "-- ", unpack(omitinfo[i]))
-- end
local cur_omit_list = {}
local cur_min_omit_x = -1
for i = 1, q do
  for j = 1, #omitinfo[i] do
    local val = omitinfo[i][j]
    if(0 < val) then
      cur_omit_list[val] = true
      if(cur_min_omit_x < 0 or val < cur_min_omit_x) then
        cur_min_omit_x = val
      end
    end
  end
  print(cur_min_omit_x)

  for j = 1, #omitinfo[i] do
    local val = omitinfo[i][j]
    if(val < 0) then
      cur_omit_list[-val] = nil
      if(cur_min_omit_x == -val) then
        cur_min_omit_x = -1
        for key, value in pairs(cur_omit_list) do
          if(cur_min_omit_x < 0 or key < cur_min_omit_x) then
            cur_min_omit_x = key
          end
        end
      end
    end
  end
end