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 Java (OpenJDK 11.0.6) (judge verdict: Time Limit Exceeded)

import java.util.*;

public class Main {

    private static class Node {
        int x;
        long v;
        Node(int x, long v) {
            this.x = x;
            this.v = v;
        }
    }

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        int m = scan.nextInt();
        TreeMap<Integer, Long> M = new TreeMap<>();
        TreeSet<Node> S = new TreeSet<>((a, b) -> {
            if (a.v < b.v || (a.v == b.v && a.x < b.x)) return -1;
            if (a.v == b.v && a.x == b.x) return 0;
            return 1;
        });
        for (int i = 1; i <= m; i++) {
            M.put(i, 0L);
            S.add(new Node(i, 0L));
        }
        for (int i = 1; i <= n; i++) {
            int a = scan.nextInt();
            int b = scan.nextInt();
            Integer fb = M.floorKey(b + 1);
            if (fb != null) {
                int x = fb.intValue();
                long tem = M.get(x) + b + 1 - x;
                if (x != b + 1 && b + 1 <= m) {
                    M.put(b + 1, tem);
                    S.add(new Node(b + 1, tem));
                }
            }
            while (true) {
                fb = M.floorKey(b);
                if (fb == null) break;
                int x = fb.intValue();
                long v = M.get(x);
                if (x < a) break;
                M.remove(x);
                S.remove(new Node(x, v));
            }
            System.out.println(S.size() == 0 ? -1 : S.first().v + i);
        }
    }
}