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;
typedef long long ll;
typedef unsigned long long ull;
#define endl '\n'
#define all(x) (x).begin(),(x).end()
const int INF=1000000000+5;
const int N=2e5+5;
const ll oo=1e18+5;
const ll mod=1e9+7;

int main(){
	ios::sync_with_stdio(0); cin.tie(0);
	int n,m;
	cin>>n>>m;
	map<int,int> maps;
	multiset<int> val;
	for(int i=0;i<m;++i) maps[i]=i;
	for(int i=0;i<m;++i) val.insert(0);
	for(int i=0;i<n;++i){
		int a,b;
		cin>>a>>b;
		--a;
		auto it=maps.lower_bound(a);
		int something=-1;
		while(it!=maps.end()&&it->first<=b){
			something=max(something,it->second);
			int diff=it->first-it->second;
			maps.erase(it++);
			val.erase(val.find(diff));
		}
		if(something>=0&&b<m){
			val.insert(b-something);
			maps[b]=something;
		}
		//~ for(int j=0;j<m;++j) if(maps.find(j)!=maps.end())cout<<"maps "<<j<<" = "<<maps[j]<<" ";
		//~ cout<<endl;
		if(val.empty()){
			cout<<-1<<endl;
			continue;
		}
		int ans=*val.begin()+i+1;
		cout<<ans<<endl;
	}		
	return 0;
}