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 INF=1e9;

int main()
{
	//1.输入 
	int n,m;
	scanf("%d%d",&n,&m);
	set<pair<int,int> > se;//记录起点,记录横向移动距离 
	multiset<int> ans;//竖直移动距离 
	//set<pair<int,int> >::iterator it; 
	//2.起点初始化 
	for(int i=1;i<=m;++i)
	{
		se.insert({i,i});
		ans.insert({0});
	}
	//3.
	for(int i=1;i<=n;++i)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		if(y == m) y=INF;
		auto it=se.lower_bound({x,0});
		int t=-1;
		for( ;it != se.end() && it->first<=y;)
		{
			ans.erase(ans.find(it->first - it->second));
			t=max(t,it->second);
			it=se.erase(it);//删除当前迭代器指向的元素并指向下一个 
		}
		if(t>-1)
		{
			ans.insert(y+1-t);
			se.insert({y+1,t});
		}
		int r= *ans.begin();
		if(r>m) cout<<"-1\n";
		else printf("%d\n",r+i);
	} 
	return 0;
}