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 PI 3.141592653589793238462
using namespace std;
typedef long long ll;
typedef long double db;
int a[200005],d[200005],b[200005];
int main(){
    int n,m;cin>>n>>m;
    for(int i=1;i<=n;i++){
        cin>>a[i]>>b[i];
    }
    set<int> s;multiset<int> dis;
    for(int i=1;i<=m;i++){
        s.insert(i);dis.insert(0);d[i]=0;
    }
    for(int i=1;i<=n;i++){
        int mi=1e9;
        auto it=s.lower_bound(a[i]);
        while(it!=s.end()&&*it<=b[i]+1){
            int x=*it;s.erase(it);
            it=s.lower_bound(a[i]);
            dis.erase(dis.lower_bound(d[x]));
            mi=min(mi,d[x]+b[i]+1-x);d[x]=1e9;
        }
        if(b[i]+1<=m&&mi<1e9){
            d[b[i]+1]=min(d[b[i]+1],mi);
            dis.insert(d[b[i]+1]);s.insert(b[i]+1);
        }
        if(dis.size()==0) cout<<-1<<endl;
        else cout<<*dis.begin()+i<<endl;
    }
}