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 <cstdio>
#include <algorithm>
#include <cstring>
#include <set>
using namespace std;
const int maxw=200050;
const int INF=1e9;
struct node{
	int pos,val;
	node(int pos=0,int val=0):pos(pos),val(val){}
};
bool operator < (node a,node b){
	return a.pos<b.pos||(a.pos==b.pos&&a.val<b.val);
}
set<node> s1;
multiset<int> s2;
node nd[maxw];
int h,w;

void solve(){
	scanf("%d %d",&h,&w);
	for(int i=1;i<=w;i++){
		s1.insert(node(i,0));
		s2.insert(0);
	}
	for(int i=1;i<=h;i++){
		int a,b;
		scanf("%d %d",&a,&b);
		auto it=s1.lower_bound(node(a,0));
		int min_v=INF,tot=0;
		while(it!=s1.end()&&(it->pos)<=b+1){
			nd[++tot]=*it;
			min_v=min(min_v,nd[tot].val+(b+1)-nd[tot].pos);
			it++;
		}
		for(int i=1;i<=tot;i++){
			s1.erase(nd[i]);
			auto it=s2.find(nd[i].val);
			s2.erase(it);
		}
		if(tot>0&&b+1<=w){
			s1.insert(node(b+1,min_v));
			s2.insert(min_v);
		}
		if(s2.empty())	printf("-1\n");
		else	printf("%d\n",i+*s2.begin());
	}
}
int main(){
	solve();
	return 0;
}