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;
#define foreach(it,v) for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define chmax(a,b) a=max(a,b)
int n,m;
map<int,int> mp;
multiset<int> v;
int main(){
	cin>>n>>m;
	for(int i=0;i<m;++i){
		mp[i]=i;
	}
	for(int i=0;i<m;++i){
		v.insert(0);
	}
	for(int i=0;i<n;++i){
		int l,r;
		cin>>l>>r;
		l--;
		auto it=mp.lower_bound(l);
		int rr=-1;
		while(it!=mp.end()&&it->first<=r){
			rr=max(rr,it->second);
			int len=it->first-it->second;
			v.erase(v.find(len));
			mp.erase(it++);
		}
		if(rr!=-1&&r<m){
			v.insert(r-rr);
			mp[r]=rr;
		}
		int ans=-1;
		if(!v.empty()){
			ans=*v.begin()+(i+1);
		}
		cout<<ans<<'\n';
	}
	return 0;
}