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<iostream>
#include<string>
#include<iomanip>
#include<cmath>
#include<vector>
#include<algorithm>
#include<utility>

using namespace std;

#define int long long
#define endl "\n"

constexpr long long INF = (long long)1e18;
constexpr long long MOD = 1'000'000'007; 

struct fast_io {
	fast_io(){
		std::cin.tie(nullptr);
		std::ios::sync_with_stdio(false);
	};
} fio;

#define A first
#define B second

signed main(){
	cout<<fixed<<setprecision(10);
	
    int H, W;
    vector<pair<int,int>> in;
    
    cin>>H>>W;
    
    in.resize(H);
    
	for(int i = 0; i < H; i++){
        cin>>in[i].A>>in[i].B;
    }
    
    int now = 1;
    bool flag = false;
    
    for(int i = 0; i < H; i++){
        if(now >= in[i].A) now = max(in[i].B + 1, now);
        
        // cout<<"i = "<<i<<endl;
        if(now > W) flag = true;
        
        if(flag == true) cout<<-1<<endl;
        else cout<<i + now<<endl;
    }
	
	
	return 0;
}