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: Time Limit Exceeded)

use proconio::{input, marker::Usize1};

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

    let mut dp = vec![(0, 0); w];
    for j in 0..w {
        dp[j] = (j, j);
    }

    for i in 1..=h {
        let e = if ab[i - 1].1 + 1 < w { ab[i - 1].1 + 1 } else { usize::max_value() };
        let l = dp.binary_search_by(|&j| j.1.cmp(&ab[i - 1].0));
        let r = dp.binary_search_by(|&j| j.1.cmp(&ab[i - 1].1));
        if let Ok(l) = l {
            if let Ok(r) = r {
                dp[r].1 = e;
                if l < r {
                    dp.drain(l..r);
                }
            } else {
                let r = r.unwrap_err() - 1;
                dp[r].1 = dp[r].1.max(e);
                if l < r {
                    dp.drain(l..r);
                }
            }
        } else {
            let l = l.unwrap_err();
            if let Ok(r) = r {
                dp[r].1 = dp[r].1.max(e);
                if l < r {
                    dp.drain(l..r);
                }
            } else {
                let r = r.unwrap_err();
                dp[r].1 = dp[r].1.max(e);
                if l < r {
                    dp.drain(l..r);
                }
            }
        }
        dbg!(&dp);
        let mut m = usize::max_value();
        for &j in &dp {
            m = m.min(j.1 - j.0);
        }
        if m < w {
            println!("{}", m + i);
        } else {
            println!("-1");
        }
    }
}