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 int ll;
typedef pair<ll,ll> P;
typedef vector<ll> VI;
typedef vector<VI> VVI;
const ll MOD = 1000000007;
const ll INF = 1e18;
#define REP(i,n) for(int i=0;i<n;i++)
#define ALL(v) v.begin(),v.end()


int main(){
	int h, w; cin >> h >> w;
	int a, b;
	set<P> st; multiset<ll> mst;
	REP(i,w){
		st.insert({i+1,i+1});
		mst.insert(0);
	}
	REP(i,h){
		cin >> a >> b;
		auto it=st.lower_bound({a,0});
		auto last=st.end();
		vector<P> era;
		while((*it).first<=b){
			era.push_back(*it);
			it++;
			if(it==last)
				break;
		}
		ll maxl=0;
		REP(i,era.size()){
			maxl=max(maxl,era[i].second);
			st.erase(era[i]);
			mst.erase(mst.find(era[i].first-era[i].second));
		}
		if(era.size()>0){
			st.insert({(b==w?INF:b+1),maxl});
			mst.insert((b==w?INF:b+1-maxl));
		}
		ll ans=*mst.begin()+i+1;
		cout << (ans<INF?ans:-1) << endl;
	}
	return 0;
}