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;
map<int,int>m;
multiset<int>s; 
int main() {
	int H,W;
	scanf("%d%d",&H,&W);
	for(int i=1;i<=W;++i) {
		m[i]=i;
		s.insert(0); 
	}
	for(int i=1;i<=H;++i) {
		int A,B;scanf("%d%d",&A,&B);
		auto it=m.lower_bound(A);
		int mx=0;
		while(it!=m.end()&&it->first<=B) {
			mx=max(mx,it->second);
			int dis=it->first-it->second;
			s.erase(s.find(dis));
			m.erase(it++);
		}
		if(mx!=0&&B<W&&m.find(B+1)==m.end()) {
			m[B+1]=mx;
			s.insert(B+1-mx);
		}
		printf("%d\n",s.empty() ? -1:(*s.begin()+i));
	}
	return 0;
}