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 int long long
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,n) for(int i=1;i<n;i++)
#define rev(i,n) for(int i=n-1;i>=0;i--)
#define all(v) v.begin(),v.end()
#define P pair<int,int>
#define len(s) (int)s.size()
 
template<class T> inline bool chmin(T &a, T b){
	if(a>b){a=b;return true;}
	return false;
}
template<class T> inline bool chmax(T &a, T b){
	if(a<b){a=b;return true;}
	return false;
}
constexpr int mod = 1e9+7;
constexpr long long inf = 3e18;

int H,W,A[200005],B[200005];
signed main(){
	cin.tie(0);ios::sync_with_stdio(false);
	cin>>H>>W;
	set<pair<int,P>>st;
	set<P>ans;
	rep(i,W){
		st.insert({i,{i,i}});
		ans.insert({0,i});
	}
	rep(i,H){
		cin>>A[i]>>B[i];
		A[i]--;B[i]--;
		auto it=st.lower_bound({A[i],{0,0}});
		int first=inf,last=-inf;
		while(it!=st.end()&&it->first<=B[i]){
			chmin(first,it->second.first);
			chmax(last,it->second.second);
			ans.erase({it->first-it->second.second,it->second.first});
			it=st.erase(it);
		}
		if(B[i]!=W-1&&first!=inf)ans.insert({B[i]+1-last,first});
		if(B[i]!=W-1&&first!=inf)st.insert({B[i]+1,{first,last}});
		if(len(ans))cout<<(ans.begin()->first)+i+1<<"\n";
		else cout<<"-1\n";
	}
}