Problem: 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.

Example

Input

4 4
2 4
1 1
2 3
2 4


Output

1
3
6
-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

/*
提出前に確認!
- setとmultisetを間違えない
*/

int	main(void)
{
	int h, w;
	cin >> h >> w;
	vector<pair<int, int>> obstacle(h);
	for (int i = 0; i < h; i++) cin >> obstacle[i].first >> obstacle[i].second;
	multiset<int> values;
	multiset<pair<int, int>> dest_and_start;
	for (int i = 0; i < w; i++)
	{
		dest_and_start.insert({i + 1, i + 1});
		values.insert(0);
	}
	for (int i = 0; i < h; i++)
	{
		//destがAiのものから順に見ていく
		auto ptr = dest_and_start.lower_bound({obstacle[i].first, -1});
		int max_start = -1;
		while (ptr != dest_and_start.end())
		{
			if ((*ptr).first > obstacle[i].second) break ;
			max_start = max(max_start, (*ptr).second);
			values.erase(values.find((*ptr).first - (*ptr).second));
			ptr = dest_and_start.erase(ptr);
		}
		if (max_start != -1 && obstacle[i].second != w)
		{
			dest_and_start.insert({obstacle[i].second + 1, max_start});
			values.insert(obstacle[i].second + 1 - max_start);
		}
		if (values.size()) cout << *values.begin() + i + 1 << endl;
		else cout << -1 << endl;
	}
	return (0);
}