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 rep(i,n) for (int i = 0; i < n; ++i)
using ll = long long;
using P = pair<int,int>;
int main() {
   int H,W;cin>>H>>W;
   vector<P> walls;
   set<P> end_start;
   multiset<int> dist;
   rep(i,H){
      int A,B;cin>>A>>B;
      A--;B--;
      walls.push_back({A,B});
      end_start.insert({i,i});
      dist.insert(0);
   }
   rep(i,H){
      P wall=walls[i];
      auto l=end_start.lower_bound({wall.first,0});
      auto r=end_start.lower_bound({wall.second+1,0});
      P best={W,0};
      int most_left=0;
      vector<P>extr;
      for(;l!=r;l++){
         int d=l->first-l->second;
         dist.extract(d);
         P es=*l;
         extr.push_back(*l);
         es.first=wall.second+1;
         if(es.second>=most_left){
            best=es;
            most_left=es.second;
         }
      }
      for (auto &&j : extr) end_start.extract(j);
      if(best.first!=W){
         end_start.insert(best);
         dist.insert(best.first-best.second);
      }
      if(dist.empty())cout<<-1<<endl;
      else cout<<(*dist.begin()+i+1)<<endl;
   }
}