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;
const int INF=0x3F3F3F3F;
int main() {
#ifndef ONLINE_JUDGE
	freopen("in.txt","r",stdin);
#endif
	int n,m;
	cin>>n>>m;
	set<pair<int,int> > se;
	multiset<int> ans;
	for(int i=1; i<=m; ++i) {
		se.insert({i,i});
		ans.insert(0);
	}
	for(int i=1; i<=n; ++i) {
		int x,y;
		cin>>x>>y;
		if(y==m) y=INF;
		auto it=se.lower_bound({x,0});
		int t=-1;
		while(it!=se.end() && it->first<=y) {
			ans.erase(ans.find(it->first-it->second));
			t=max(t,it->second);
			it=se.erase(it);
		}
		if(t>-1) {
			ans.insert(y+1-t);
			se.insert({y+1,t});
		}
		int r=*ans.begin();
		if(r>m) cout<<"-1\n";
		else cout<<r+i<<endl;
	}
	return 0;
}