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;
int n,m,dis[202020];
set<int>st;
multiset<int>sstt;
int main()
{
	cin>>n>>m;
	for(int i=0;i<m;i++)
	{
		st.insert(i);
		sstt.insert(0);
	}
	for(int i=0;i<n;i++)
	{
		int a,b;
		cin>>a>>b;
		a--;
		set<int>::iterator it=st.lower_bound(a),it1;
		int mn=1e9+1e5;
		while(it!=st.end() && *it<=b)
		{
			int x=*it;
			mn=min(mn,dis[x]+b-x);
			it1=it;
			it1++;
			st.erase(it);
			it=it1;
			sstt.erase(sstt.find(dis[x]));
			dis[x]=1e9+1e5;
		}
		if(b!=m && mn!=1e9+1e5)
		{
			dis[b]=mn;
			st.insert(b);
			sstt.insert(mn);
		}
		if(st.empty())
		{
			cout<<"-1\n";
		}
		else
		{
			cout<<*sstt.begin()+i+1<<endl;
		}
	}
	return 0;
}