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;

int H,W,A[200000],B[200000],C[200000];
set<pair<int,int>>st;
priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>>pq;

int main(){
	cin>>H>>W;
	for(int i=0;i<H;i++){
		cin>>A[i]>>B[i];
		A[i]--;
		B[i]--;
	}
	for(int i=0;i<W;i++){
		C[i]=i;
		pq.push({0,i});
		st.insert({i,i});
	}
	for(int i=0;i<H;i++){
		auto it=st.lower_bound({A[i],-1});
		int t=-1;
		while(it!=st.end()&&it->first<=B[i]){
			t=it->second;
			C[it->second]=-1;
			auto it2=it;
			it++;
			st.erase(it2);
		}
		if(B[i]!=W-1&&t!=-1){
			C[t]=B[i]+1;
			st.insert({B[i]+1,t});
			pq.push({B[i]+1-t,t});
		}
		while(!pq.empty()&&pq.top().first+pq.top().second!=C[pq.top().second]){
			pq.pop();
		}
		if(pq.empty()){
			cout<<-1<<endl;
		}
		else{
			cout<<i+1+pq.top().first<<endl;
		}
	}
}