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<bits/stdc++.h>
  
#define lol long long
#define gcd(x,y) __gcd(x,y)
#define mt make_tuple
#define mp make_pair
#define fi first
#define se second
#define fixed(x) fixed<<setprecision(x)
using namespace std;
using pii=pair<int,int>;
template <class A, class B> inline bool chmax(A &a, const B &b) { return b > a && (a = b, true); }
template <class A, class B> inline bool chmin(A &a, const B &b) { return b < a && (a = b, true); }
template <class A> inline A abs(A &a) { return (a < 0 ? -a : a); }
bool inLine(int x,int y,int mx,int my){return (x>=0&&y>=0&&x<mx&&y<my);}
const lol inf=(1LL<<62);
const lol MOD=(1e9)+7;
const lol mod=998244353;
const int dx[]={1,0,-1,0,1,1,-1,-1};
const int dy[]={0,1,0,-1,1,-1,1,-1};
signed main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  int h,w;
  cin >>h>>w;
  vector<int> a(h),b(h);
  for(int i=0;i<h;i++){
    cin >>a[i]>>b[i];
  }
  priority_queue<tuple<int,int,int>,vector<tuple<int,int,int>>,greater<tuple<int,int,int>>> pq;
  for(int i=1;i<=w;i++){
    pq.push(mt(0,0,i));
  }
  vector<int> ans(h+3,-1);
  while(!pq.empty()){
    int cost,deep,wide;
    tie(cost,deep,wide)=pq.top();
    pq.pop();
    if(wide>w||deep>h) continue;
    if(ans[deep]==-1) ans[deep]=cost;
    if(a[deep]<=wide&&wide<=b[deep]){
      pq.push(mt(cost+1,deep,wide+1));
    }else{
      pq.push(mt(cost+1,deep+1,wide));
    }
  }
  for(int i=1;i<=h;i++) cout <<ans[i]<<'\n';
  return(0);
}