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 ll;
ll read(){
	ll a=0,b=getchar(),c=1;
	while(!isdigit(b))c=b=='-'?-1:1,b=getchar();
	while(isdigit(b))a=a*10+b-'0',b=getchar();
	return a*c;
}
map<int,int> m;
multiset<int> s;
int main(){
	int h=read(),w=read();
	for(int i=0;i<w;i++)
		m[i]=i,s.insert(0);
	for(int i=0;i<h;i++){
		int a=read()-1,b=read(),mx=-1;
		map<int,int>::iterator x=m.lower_bound(a);
		while(x!=m.end() and x->first<=b){
			mx=max(mx,x->second);
			int y=x->first-x->second;
			s.erase(s.find(y));
			m.erase(x++);
		}
		if(mx!=-1 and b<w){
			s.insert(b-mx);
			m[b]=mx;
		}
		if(s.empty())puts("-1");
		else cout << *s.begin()+i+1 << endl;
	}
	return 0;
}