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;
#define int long long
const int mxn=2e5+3,du=1e9+7;
int m,n;
signed main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>m>>n;
	map<int,int> mp;
	multiset<int> s;
	for (int i=1;i<=n;i++) mp[i]=i,s.insert(0);
	for (int i=1;i<=m;i++)
    {
        int l,r;
        cin>>l>>r;
        auto it=mp.lower_bound(l);
        int mx=0;
        while(it!=mp.end()&&it->first<=r)
        {
            mx=max(mx,it->second);
            int dis=(it->first)-(it->second);
            s.erase(s.find(dis));
            mp.erase(it++);
        }
        if (mx!=0&&r<n&&mp.find(r+1)==mp.end())
        {
            mp[r+1]=mx;
            s.insert(r+1-mx);
        }
       // cout<<"DM";
        if (s.size()) cout<<(*s.begin())+i<<"\n";
        else cout<<-1<<"\n";
    }
	return 0;
}