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 pair<long,long> P;
int main(){
  long H,W;cin>>H>>W;
  set<P> S,T;
  for(long i=0;i<W;i++){S.insert(P(i,0));T.insert(P(0,i));}
  for(long k=0;k<H;k++){
    long a,b;cin>>a>>b;a--;
    long y=LONG_MAX;
    for(auto itr=S.lower_bound(P(a,-1));itr!=S.end()&&(itr->first)<b;){
      P x=*itr;
      y=min(y, x.second+b-x.first);
      T.erase(P(x.second,x.first));
      itr=S.erase(itr);
    }// ゴール: 移動回数 i,jとして, b, j+b-i, このうち, 最も移動回数が少ないものだけ残す
    if(y!=LONG_MAX&&b<W){S.insert(P(b,y)); T.insert(P(y,b));}
    cout<<(T.size()?(T.begin()->first)+k+1:-1)<<endl;
  }
}

/*

S: ゴール, 何回の移動でいけるか
T: 何回の移動でいけるか, ゴール

T.begin()を持って来ればいける
Sで更新を素早くみる





*/