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 OCaml (4.10.0) (judge verdict: Wrong Answer)

let () = Scanf.scanf "%d %d\n" @@ fun h w ->
  let as_ = Array.make h 0 in
  let bs = Array.make h 0 in
  for i = 0 to h - 1 do
    Scanf.scanf "%d %d\n" @@ fun a b ->
      as_.(i) <- a;
      bs.(i) <- b
  done;
  let dp = Array.make (h + 1) 1 in
  for i = 0 to h - 1 do
    dp.(i + 1) <- if dp.(i) < as_.(i) then dp.(i) else bs.(i) + 1
  done;
  let dp' = Array.make (h + 1) w in
  for i = h - 1 downto 0 do
    dp'.(i) <- if bs.(i) < dp.(i + 1) then dp.(i + 1) else as_.(i) - 1
  done;
  for i = 1 to h do
    Printf.printf "%d\n" @@
    if w < dp.(i)
    then -1
    else i + min (dp.(i) - 1) (dp'.(i) - dp'.(0))
  done