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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

#define max(a, b) ((a > b) ? a : b)
#define min(a, b) ((a < b) ? a : b)

#define forn(n) for(int i = 0; i < n; i++)
#define fork(k, n) for(int i = k; i < n; i++)

#define pb(n) push_back(n);

typedef long long ll;
typedef double dbl;

const int MX = 1e6;
const int MOD = 1e9 + 7;
const string ALPHA = "abcdefghijklmnopqrstuvwxyz";
const double PI = 3.1415926535;


int main(){
  ios::sync_with_stdio(0);
  cin.tie(0);
	//Code here bby
  int h, w;
  cin>>h>>w;
  int curr = 1;
  vector<int> dists;
  dists.push_back(0);
  forn(h){
    int a, b;
    cin>>a>>b;
    bool ok = 0;
      for(int drop = curr; drop <= w; drop++){
        if(dists[dists.size() - 1] == -1){
          break;
        }
        if(drop >= a && drop <= b){
          continue;
        }
        dists.push_back(dists[dists.size() - 1] + 1 + (drop - curr));
        curr = drop;
        ok = 1;
        break;
      }
    if(!ok){
      dists.push_back(-1);
    }
  }
  for(int idx : dists){
    if(idx == 0){
      continue;
    }
    cout<<idx<<endl;
  }
}