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: Accepted)

#pragma GCC optimize ("O2")
#pragma GCC target ("avx2")
#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
#define rep(i, n) for(int i = 0; i < (n); i++)
#define rep1(i, n) for(int i = 1; i <= (n); i++)
#define co(x) cout << (x) << "\n"
#define cosp(x) cout << (x) << " "
#define ce(x) cerr << (x) << "\n"
#define cesp(x) cerr << (x) << " "
#define pb push_back
#define mp make_pair
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
#define Would
#define you
#define please


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


	int H, W;
	cin >> H >> W;

	map<int, int> MP;
	rep(i, W) {
		MP[i] = 1;
	}

	int kazu[200201] = {};
	kazu[1] = W;
	kazu[200100] = 1;
	int ima = 1;

	rep(i, H) {
		int a, b;
		cin >> a >> b;
		a--;
		b--;

		auto s = MP.lower_bound(a);
		auto t = MP.upper_bound(b);

		int saishou = 1e9;
		for (; s != t; s++) {
			chmin(saishou, (*s).second + b - (*s).first + 1);
			kazu[(*s).second]--;
		}
		MP.erase(MP.lower_bound(a), t);

		b++;
		if (saishou < 1e8) {
			if (MP.find(b) != MP.end()) {
				if (b != W) {
					int tmp = MP[b];
					kazu[tmp]--;
					MP[b] = min(tmp, saishou);
					kazu[min(tmp, saishou)]++;
				}
			}
			else {
				if (b != W) {
					MP[b] = saishou;
					kazu[saishou]++;
				}
			}
		}

		while (kazu[ima] == 0) ima++;
		int ans = i + ima;
		if (ima > 200050) co(-1);
		else co(ans);
	}





	Would you please return 0;
}