Problem:
Score : 600 points

Problem Statement

There is a grid of squares with H+1 horizontal rows and W vertical columns.

You will start at one of the squares in the top row and repeat moving one square right or down. However, for each integer i from 1 through H, you cannot move down from the A_i-th, (A_i + 1)-th, \ldots, B_i-th squares from the left in the i-th row from the top.

For each integer k from 1 through H, find the minimum number of moves needed to reach one of the squares in the (k+1)-th row from the top. (The starting square can be chosen individually for each case.) If, starting from any square in the top row, none of the squares in the (k+1)-th row can be reached, print -1 instead.

Constraints

1 \leq H,W \leq 2\times 10^5

1 \leq A_i \leq B_i \leq W

Input

Input is given from Standard Input in the following format:

H W
A_1 B_1
A_2 B_2
:
A_H B_H

Output

Print H lines. The i-th line should contain the answer for the case k=i.

Sample Input 1

4 4
2 4
1 1
2 3
2 4

Sample Output 1

1
3
6
-1

Let (i,j) denote the square at the i-th row from the top and j-th column from the left.

For k=1, we need one move such as (1,1) → (2,1).

For k=2, we need three moves such as (1,1) → (2,1) → (2,2) → (3,2).

For k=3, we need six moves such as (1,1) → (2,1) → (2,2) → (3,2) → (3,3) → (3,4) → (4,4).

For k=4, it is impossible to reach any square in the fifth row from the top.

Solution in Kotlin (1.3.71) (judge verdict: Time Limit Exceeded)

import java.util.*

fun main(args : Array<String>) {
    val (h, w) = readLine()!!.split(" ").map { it.toInt() }
    val blocks = mutableMapOf<Pair<Int, Int>, Boolean>()

    repeat(h) {
        val (a, b) = readLine()!!.split(" ").map { it.toInt()-1 }
        for (i in a..b) {
            blocks[Pair(it, i)] = true
        }
    }

    val bfs = PriorityQueue<Step> { s1, s2 ->
        if (s1.steps == s2.steps) {
            return@PriorityQueue s2.hIdx - s1.hIdx
        } else {
            return@PriorityQueue s1.steps - s2.steps
        }
    }
    val seen = mutableMapOf<Pair<Int, Int>, Int>()

    repeat(w) {
        bfs.add(Step(0, it, 0))
        seen[Pair(0, it)] = 0
    }

    val ans = BooleanArray(h+1){false}
    var maxHIdx = 0

    while (bfs.isNotEmpty()) {
        val step = bfs.poll()

        if (step.hIdx > 0) {
            if (!ans[step.hIdx]) {
                println(step.steps)
                ans[step.hIdx] = true
                maxHIdx = step.hIdx

                if (step.hIdx == h) return
            }
        }

        if (step.hIdx < h &&
            blocks[Pair(step.hIdx, step.wIdx)] != true &&
            (seen[Pair(step.hIdx+1, step.wIdx)] ?: Int.MAX_VALUE) > step.steps+1) {
            bfs.add(Step(step.hIdx+1, step.wIdx, step.steps+1))
            seen[Pair(step.hIdx+1, step.wIdx)] = step.steps+1
        } else if (step.wIdx != w-1 &&
            (seen[Pair(step.hIdx, step.wIdx+1)] ?: Int.MAX_VALUE) > step.steps+1) {
            bfs.add(Step(step.hIdx, step.wIdx+1, step.steps+1))
            seen[Pair(step.hIdx, step.wIdx+1)] = step.steps+1
        }
    }

    repeat(h - maxHIdx) {
        println(-1)
    }
}

data class Step(val hIdx: Int, val wIdx: Int, val steps: Int)