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 <map>
#include <set>
#include <ctime>
#include <cmath>
#include <stack>
#include <queue>
#include <bitset>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring> 
#include <iostream>
#include <algorithm>
using namespace std;
 
int H, W;
int l, r;
set< pair<int, int> > s;
multiset<int> ms;
 
int main()
{
	scanf("%d%d", &H, &W);
	for(int i = 1;i <= W;++i)
		s.insert(make_pair(i, i)), ms.insert(0);
 
 
	for(int i = 1;i <= H;++i)
	{
		int l, r;
		scanf("%d%d", &l, &r);
 
		set< pair<int, int> >::iterator it;
		int mx = -1, xx = 0;
		vector< pair<int, int> > v;
		for(it = s.lower_bound(make_pair(l, 0));it != s.end() && (*it).first <= r;it++)
		{
			//cerr << (*it).first << " " << (*it).second << endl;
			mx = max(mx, (*it).second);
			int num = (*it).first - (*it).second;
			ms.erase(ms.find(num));
			v.push_back(*it);
		}
		int si = v.size();
		for(int j = 0;j < si;++j)
		{
			pair<int, int> a = v[j];
			s.erase(s.find(a));
		}
		if(mx != -1 && r + 1 <= W)
		{
			ms.insert(r + 1 - mx);
			s.insert(make_pair(r + 1, mx));
		}
		if(ms.size() > 0)
			printf("%d\n", *ms.begin() + i);
		else
			printf("-1\n");
	}
 
	return 0;
}