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:
//hail to jwalaji
#include<bits/stdc++.h>
using namespace std;
typedef long long int lli;
lli h,w,a,b,f,t,e;
set<pair<lli,lli>> st;
multiset<lli> ms;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cin>>h>>w;
    for(lli i=0;i<w;i++){
        st.insert({i,i});
        ms.insert(0);
    }
    for(lli i=0;i<h;i++){
        cin>>a>>b;
        auto tl=st.lower_bound({a-1,0}),el=st.lower_bound({b,0});
       e=tl!=el;
       for(;tl!=el;tl=st.erase(tl)){
        tie(t,f)=*tl;
        ms.erase(ms.find(t-f));
       }
       if(e&&b!=w){
        st.insert({b,f});
        ms.insert(b-f);
       }
       if(ms.empty()){cout<<-1<<endl;}
       else{
        cout<<*ms.begin()+i+1<<endl;
       }
    }
}