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 N=2e5+5;
multiset<int>val;
map<int,int>mp;
int main(){
	int h,w,l,r;scanf("%d%d",&h,&w);
	for(int i=1;i<=w;i++)val.insert(0),mp[i]=0;
	for(int i=1;i<=h;i++){
		scanf("%d%d",&l,&r);
		if(r!=w){
			auto x=mp.lower_bound(r+1);
			if(x->first!=r+1&&x!=mp.begin()){
				x--;
				int v=x->second+(r+1-x->first);
				mp[r+1]=v,val.insert(v);
			}
		}
		auto x=mp.lower_bound(l);
		while(x!=mp.end()&&x->first<=r)val.erase(val.find(x->second)),x=mp.erase(x);
		if(!val.empty())printf("%d\n",i+*val.begin());
		else puts("-1");
	}
	return 0;
}