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;
int n,m;
map<int,int>mp;
multiset<int>val;
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)mp[i]=0,val.insert(0);
	for(int i=1;i<=n;i++){
		int L,R;scanf("%d%d",&L,&R);
		if(R!=m){
			auto it=mp.upper_bound(R+1);
			if(it!=mp.begin()){
				it--;
				if(it->first!=R+1){
					int v=it->second+(R+1-it->first);
					mp[R+1]=v;
					val.insert(v);
				}
			}
		}
		auto it=mp.lower_bound(L);
		while(it!=mp.end()&&it->first<=R){
			val.erase(val.find(it->second));
			it=mp.erase(it);
		}
		if(!val.empty())printf("%d\n",i+*val.begin());
		else puts("-1");
	}
}