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: Wrong Answer)

import java.util.*

fun main() {
    val pw = java.io.PrintWriter(System.out)
    Problem.solve(Scanner(System.`in`), pw)
    pw.flush()
}

object Problem {
    fun solve(sc: Scanner, pw: java.io.PrintWriter) {
        val H = sc.next().toInt()
        val W = sc.next().toInt()
        var x = 0
        var s = 0
        repeat(H) {
            val A = sc.next().toInt() - 1
            val B = sc.next().toInt() - 1
            if (x >= W) {
                pw.println("-1")
            } else {
                s++
                if (x >= A) {
                    s += (B + 1 - x)
                    x = B + 1
                }
                pw.println(if (x < W) s else "-1")
            }
        }
    }
}