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;

int h,w;
pair<int,int> v[200001];
int res[200005];

int main(void)
{
	cin.tie(0);
	ios::sync_with_stdio(false);

	cin >> h >> w;

	for(int i=1;i<=h;i++)
	{
		cin >> v[i].first >> v[i].second;
		res[i] = 1e9;
	}
	res[h+1] = 1e9;

	if(v[1].first > 1)
	{
		int p = 1;
		int cnt = 0;
		for(int i=2;i<=h+1;i++)
		{
			cnt++;
			res[i] = min(res[i],cnt);
			if(i==h+1)
			{
				break;
			}
			if(p < v[i].first || v[i].second < p)
			{
				continue;
			}
			else
			{
				int d = v[i].second - p + 1;
				cnt+=d;
				p = v[i].second + 1;
				if(p > w)
				{
					break;
				}
			}
		}
	}
	if(v[1].second < w)
	{
		int p = v[1].second + 1;
		int cnt = 0;
		for(int i=2;i<=h+1;i++)
		{
			cnt++;
			res[i] = min(res[i],cnt);
			if(i==h+1)
			{
				break;
			}
			if(p < v[i].first || v[i].second < p)
			{
				continue;
			}
			else
			{
				int d = v[i].second - p + 1;
				cnt+=d;
				p = v[i].second + 1;
				if(p > w)
				{
					break;
				}
			}
		}
	}

	for(int i=2;i<=h+1;i++)
	{
		if(res[i]==1e9)
		{
			res[i] = -1;
		}
		cout << res[i] << '\n';
	}
	return 0;	
}