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![i32::max_value(); w];
    for j in 0..w {
        dp[j] = 0;
    }
    for i in 1..=h {
        let mut v = if dp[0] < i32::max_value() && ab[i - 1].0 > 0  {
            dp[0] + 1
        } else {
            i32::max_value()
        };
        dp[0] = v;
        let mut min = v;
        for j in 1..w {
            if v < i32::max_value() {
                v += 1;
            }
            if dp[j] < i32::max_value() && j < ab[i - 1].0 || j > ab[i - 1].1 {
                v = v.min(dp[j] + 1);
            }
            dp[j] = v;
            min = min.min(v);
        }
        println!("{}", if min == i32::max_value() {
            -1
        } else {
            min
        });
    }
}