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>
#define ll long long
#define ull unsigned long long
#define db double
#define ld long double
#define pii pair<int,int>
#define vvi vector<vector<int> >
#define vpi vector<pii>
#define F first
#define S second
#define pb push_back
#define INF 0x7fffffff
#define LINF 0x7fffffffffffffff
#define MOD 1000000007
#define mod 998244353
#define lowbit(x) x&(-x)
#define HS 173LL
using namespace std;
int read()
{
	int x=0;
	char ch=getchar();
	while(!isdigit(ch))
		ch=getchar();
	while(isdigit(ch))
	{
		x=x*10+ch-'0';
		ch=getchar();
	}
	return x;
}
int h,w;
map<int,int> mp;
multiset<int> st;
int main()
{
	h=read(),w=read();
	for(int i=1;i<=w;i++)
		mp[i]=0;
	for(int i=1;i<=w;i++)
		st.insert(0);
	for(int i=1;i<=h;i++)
	{
		int l=read(),r=read();
		if(r!=w)
		{
			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;
					st.insert(v);
				}
			}
		}
		auto it=mp.lower_bound(l);
		while(it!=mp.end()&&it->first<=r)
		{
			st.erase(st.find(it->second));
			it=mp.erase(it);
		}
		if(!st.empty())
			cout<<*st.begin()+i<<endl;
		else
			cout<<-1<<endl;
	}
	return 0;
}