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;
typedef long long ll;
const int maxn=2e5+5;
const int INF=1e9+5;
set<int> se;
multiset<int> mse;
int dis[maxn]={0};///到第i位置的最小值
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0),cout.tie(0);
    int h,w;
    cin>>h>>w;
    for(int i=0;i<w;i++)
    {
        se.insert(i);///存目前可以到达的点 （0-w-1）
        mse.insert(0);///存到到达点的距离
    }
    for(int i=0;i<h;i++)
    {
        int l,r;
        cin>>l>>r;
        l--;
        int Min=INF;
        set<int>::iterator it=se.lower_bound(l);
        while(it!=se.end()&&*it<=r)
        {
            int x=*it;
            //cout<<x<<endl;
            Min=min(Min,dis[x]+r-x);
            it=se.erase(it);
            mse.erase(mse.find(dis[x]));
            dis[x]=INF;
        }
        if(r!=w&&Min!=INF)
        {
            dis[r]=min(Min,dis[r]);
            se.insert(r);
            mse.insert(dis[r]);
        }
        if(!mse.empty())
        {
            cout<<*mse.begin()+i+1<<"\n";
        }
        else
        {
            cout<<"-1"<<"\n";
        }

    }
    return 0;
}