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 Kotlin (1.0.0) (judge verdict: Accepted)

import java.util.*

private data class Work(val start: Int, val end: Int, val dist: Int)
private data class Position(val pos: Int, val index: Int)

fun main(args: Array<String>) {

    /**
     * 同等のデータ構造, ロジックを実装しているが TLE で通らない,, java と kotlin の速度差?? どう分析するべきか
     * WAをあえて出すと実行時間を出力できる
     *
     * -> !! 最後の出力形式を for 文から　joinToString に変えると一気に早くなってAC
     */
    val (n, q) = readLine()!!.split(" ").map { it.toInt() }

    // 工事を格納 O(N)
    var workList = mutableListOf<Work>()

    for (i in 0 until n) {
        val (s, t, x) = readLine()!!.split(" ").map { it.toInt() }

        // 切り上げ，切り捨てをして閉区間で考える
        var start = Math.ceil(s - 0.5 - x).toInt()
        var end = Math.floor(t - 0.5 - x).toInt()

        workList.add(Work(start, end, x))
    }

    // 工事を距離が近い順にソート O(NlogN)
    workList.sortBy { it.dist }

    // スタート地点を配列に保持する（最初からソートされて渡されるのでソートの必要なし） O(N)
    // スタート地点の中で，ある工事区間に含まれるものを取り出し，配列から削除,
    var posSet = TreeSet<Position> { a, b -> a.pos - b.pos } // Comparator をラムダ式で渡す
    for (i in 0 until q) {
        posSet.add(Position(readLine()!!.toInt(), i))
    }


    // 答えにはデフォルトで -1 を設定 (選ばれなかったものは工事に合わない = -1 )
    val ans = IntArray(q) { -1 }


    for (work in workList) {

        // work.start よりも大きな開始地点を持つ posList 内の 最初の要素を指す イテレータを取得
        // O(logN)
        val comp = Position(work.start, -1)
        var position = posSet.ceiling(comp)

        // 工事に含まれる位置にいる posList の要素を取得し，削除
        while (position != null) {
            if (position.pos > work.end) {
                break
            }

            // O(logN)
            posSet.remove(position)

            // O(1)
            ans[position.index] = work.dist

            position = posSet.ceiling(comp)
        }
    }

    println(ans.joinToString("\n"))
//    for (a in ans) {
//        println(a)
//    }
}