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 int long long
using namespace std;
using P=pair<int,int>;
signed main(){
  int H,W;
  cin>>H>>W;
  set<P> s;
  multiset<int> ans;
  for(int i=0;i<W;i++){
    s.insert({i,i}); 
    ans.insert(0);
  }
  int n=0;
  for(int i=0;i<H;i++){
    int a,b;
    cin>>a>>b;
    a--;
    int mx=-1;
    auto it=s.lower_bound({max(a,n),0});
    for(;it!=s.end() && (*it).first<=b;it=s.erase(it)){
      mx=max(mx,(*it).second);
      ans.erase(ans.find((*it).first-(*it).second));
    }
    if(b!=W){
      if(mx!=-1){
        s.insert({b,mx});
        ans.insert(b-mx);
      }
    }
    home:;
    printf("%d\n",ans.size()==0?-1:*ans.begin()+i+1);
    if(n>=a)n=max(n,b);
  }
}