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 Rust (1.42.0) (judge verdict: Wrong Answer)

#![allow(clippy::needless_range_loop)]
#![allow(unused_macros)]
#![allow(dead_code)]
#![allow(unused_imports)]

use itertools::Itertools;
use proconio::input;
use proconio::marker::*;

fn main() {
    input! {
        h: usize,
        w: usize,
        ob: [(Usize1, Usize1); h]
    }

    let mut costs = vec![100000000000usize; h];
    let mut left = vec![100000000000usize; h];

    for x in 0..w {
        for y in 0..h {
            let pl = if y == 0 {
                x
            } else {
                left[y - 1]
            };
    
            left[y] = if pl < ob[y].0 || ob[y].1 < pl {
                pl
            } else if ob[y].1 < (w - 1) {
                ob[y].1 + 1
            } else {
                break;
            };

            let cost = left[y] - x + y + 1;

            if cost < costs[y] {
                costs[y] = cost;
            } else {
                break;
            }
        }
    }

    for (i, &l) in costs.iter().enumerate() {
        if l != 100000000000usize {
            println!("{}", l);
        } else {
            println!("-1");
        }
    }
}