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 AC 0
int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	int h,w;
	cin>>h>>w;
	map<int,int> val;
	multiset<int> ans;
	for(int i=0;i<w;i++) {
		val.insert(pair<int,int>(i,0));
		ans.insert(0);
	}
	for(int i=0,a,b;i<h;i++) {
		cin>>a>>b;
		a--;
		int mn=2e9;
		while(true) {
			map<int,int>::iterator mapit=val.lower_bound(a);
			if(mapit==val.end()||mapit->first>=b) break;
			ans.erase(ans.lower_bound(mapit->second));
			mn=min(mn,mapit->second+b-mapit->first);
			val.erase(mapit);
		}
		if(mn<2e9&&b<w) {
			if(val.count(b)) {
				ans.erase(ans.lower_bound(val[b]));
				val[b]=min(val[b],mn);
			}
			else val[b]=mn;
			ans.insert(val[b]);
		}
		if(ans.size()) cout<<*ans.begin()+i+1<<'\n';
		else cout<<"-1\n";
	}
	return AC;
}