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: Runtime Error)

#include <bits/stdc++.h>

using namespace std;

#define FOR(i, n) for(int i = 0; i < (int) n; i++)
#define PB push_back
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define F first
#define S second

typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pii;

const int MOD = 1e9+7;

int moves[200007][2];

bool isOut(int i, int j){
	return j+1 < moves[i][0] || moves[i][1] < j+1;
}

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

	int h, w; cin >> h >> w;

	vector<vi> grid(h+1, vi(w));
	FOR(i, h) cin >> moves[i][0] >> moves[i][1];
	
	bool ok = true;
	FOR(i, h){
		if(!ok){
			cout << "-1\n";
			continue;
		}

		int ans = 1e9;
		if(i){
			FOR(j, w){
				if(j > 0 && grid[i][j] == 0 && grid[i][j-1] != 0){
					grid[i][j] = grid[i][j-1]+1;
				}
				if(grid[i][j] && isOut(i, j)){
					ans = min(ans, grid[i][j]+1);
					grid[i+1][j] = grid[i][j]+1;
				}
			}
		}else{
			FOR(j, w){
				if(isOut(i, j)){
					grid[i+1][j] = 1;
					ans = 1;
				}
			}
		}

		if(ans == 1e9){
			ok = false;
			cout << "-1\n";
		}else{
			cout << ans << '\n';
		}
	}


	return 0;
}