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 ll = int_fast64_t;
using P = std::pair<ll,ll>;

#define REP(i,b,e) for(ll i=b; i<e; i++)
#define ALL(vec) vec.begin(),vec.end()
#define PRINT(vec) printf("[ ");for(auto &i:vec)printf("%ld ",i);puts("]");
#define fi first
#define se second

const int MOD = 1e9+7;
const ll INF = 1e18;
int dx[] = {0, 1, 0, -1, 1, 1, -1, -1}, dy[] = {1, 0, -1, 0, 1, -1, -1, 1};

int main(){
	int h, w;
	scanf("%d %d", &h, &w);
	

	std::vector<P> ab;
	int l = 1, cnt = 0, cango = 0;
	REP(i, 0, h){
		int a, b;
		scanf("%d %d", &a, &b);
		if(l<=w){
			cango = i;
			ab.emplace_back(a, b);
			continue;
		}
		if(l<a) cnt++;
		else {
			if(b!=w) l = b+1;
			else l = w+1;
		}
		ab.emplace_back(a, b);
	}

	int r = w;
	for(int i=cango; i>=0; i--){
		auto [a, b] = ab[i];
		if(b>=r){
			if(a!=1) r = a-1;
		}
	}
	
	l = r, cnt = 0;
	REP(i, 0, h){
		auto [a, b] = ab[i];
		if(l>w){
			puts("-1");
			continue;
		}

		if(l<a) cnt++;
		else {
			if(b!=w){
				cnt += b-l+2;
				l = b+1;
			}else{
			       l = w+1;
			       puts("-1");
			       continue;
			}
		}
		printf("%d\n", cnt);
	}
	return 0;
}