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 C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <iostream>
#include <stdio.h>
#include <vector>
#include <set>
#include <string.h>
using namespace std;

int main()
{
  int h = 0;
  int w = 0;
  int l = 0;
  int u = 0;
  scanf("%d %d", &h, &w);
  long long max_value = (h+1) * (w+1);
  vector<long long> a(w, 0);
  long long cur_min = 0;
  for(int i=0; i<h; i++){
    //for(int j=0; j<w; j++){
    //  printf("%lld ", a[j]);
    //}
    //printf("\n");
    scanf("%d %d", &l, &u);
    if( l > 1 ){
      a[0] = std::min(a[0] + 1, max_value);
    }else{
      a[0] = max_value;
    }
    cur_min = a[0];
    for(int j=1; j<w; j++){
      if( j>=l-1 & j<=u-1 ){
        a[j] = std::min(a[j-1] + 1, max_value);
      }else{
        a[j] = std::min(std::min(a[j-1], a[j]) + 1, max_value);
      }
      cur_min = std::min(a[j], cur_min);
    }
    if(cur_min == max_value){
      cur_min = -1;
    }
    printf("%lld\n", cur_min);

  }
}