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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std ;
#define int long long
#define fi first
#define se second
#define pb push_back
#define pii pair<int, int>
#define mod 1000000007

typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;

int h,w ;
int arr[500005] ;
string str ;

void solve()
{
    cin >> h >> w ;
    int pos = 1 ;
    int ans = 0 ;
    int f = 0 ;
    for(int i=1 ; i<=h ; i++)
    {
        int l,r ;
        cin >> l >> r ;
        if(f)
        {
            cout << -1 << "\n" ;
            continue ;
        }
        if(pos>=l&&pos<=r)
        {
            if(r==w)
            {
                pos = w+1 ;
                cout << -1 << "\n" ;
                f = 1 ;
            }
            else
            {
                ans += (r+1-pos) ;
                pos = r+1 ;
                cout << ans+1 << "\n" ;
                ans++ ;
            }
        }
        else
        {
            cout << ans+1 << "\n" ;
            ans++ ;
        }
    }
}
int32_t main()
{
	ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
	int tt = 1;
	//cin >> tt ;
	while(tt--)
	{
	     solve() ;
	}
}