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 Haskell (GHC 7.10.3) (judge verdict: Wrong Answer)

import Control.Arrow
import Control.Monad
import Data.List
import qualified Data.ByteString.Char8 as B
import qualified Data.Vector as V
import Data.Function
import Data.Ord
--import Debug.Trace

main :: IO ()
main = readLine >>= main'
    where
        main' [n, q] = inputs >>= printRes . uncurry solve  
            where
                inputs = (,) <$> readContents n <*> replicateM q (head <$> readLine) 
        printRes = mapM_ print 

solve :: [[Int]] -> [Int] -> [Int]
solve stxs = map check 
    where
        check d = val . edit $ binarySearch (E (d, inf)) vs
            where
                edit (Right i) = vs V.! i
                edit (Left i ) = vs V.! (i - 1)

        vs = V.fromList . map mf . {-traceShowId . build'-} build . sortBy cmpf $ [-inf, inf, inf] : map mf2 stxs
            where
                mf2 [s, t, x] = [s - x, t - x, x]

                cmpf lhs rhs = comparing head lhs rhs `mappend` comparing last lhs rhs

                mf (s, x)
                    | x < inf = E (s, x)
                    | otherwise = E (s, -1)

                build (x@[s0, t0, x0] : y@[s1, t1, x1] : zs)
                    | t0 < s1 = (s0, x0) : build (ordf $ [t0, s1, inf] : y : zs)
                    | t0 == s1 = (s0, x0) : build (y : zs)
                    | t0 < t1 && x0 < x1 = (s0, x0) : build (ordf $ [t0, t1, x1] : zs)
                    | t0 < t1 = (s0, x0) : build (y : zs)
                    | x0 < x1 = build $ x : zs
                    | otherwise = (s0, x0) : build (ordf $ [t1, t0, x0] : y : zs)
                build [[s, _, x]] = [(s, x)]
                build _ = []

                --build' = build . traceShowId

                ordf (x@[s0, _, x0] : y@[s1, _, x1] : zs)
                    | s0 < s1 = x : y : zs
                    | s0 > s1 || x0 > x1 = y : ordf (x : zs)
                    | otherwise = x : y : zs
                ordf xs = xs

        inf = 10 ^ 10

        val (E (_, v)) = v

newtype E = E (Int, Int) 
    deriving (Eq)

instance Ord E where
    compare (E lhs) (E rhs) = compare lhs rhs

readLine :: IO [Int]
readLine = unfoldr (fmap (second (maybe B.empty snd . B.uncons)) . B.readInt) <$> B.getLine

readContents :: Int -> IO [[Int]]
readContents = flip replicateM readLine

binarySearch :: (Ord a) => a -> V.Vector a -> Either Int Int
binarySearch k xs
    | V.null xs = Left 0
    | otherwise = bsf 0 $ V.length xs
    where
        bsf base size
            | size > 1 && mv > k = bsf base $ size - half
            | size > 1 = bsf mid $ size - half
            | bv == k = Right base
            | bv < k = Left $ base + 1
            | otherwise = Left base
            where
                half = size `div` 2
                mid = base + half
                bv = xs V.! base
                mv = xs V.! mid