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;

using Int = long long;
const char newl = '\n';

//INSERT ABOVE HERE
signed main(){
  cin.tie(0);
  ios::sync_with_stdio(0);

  int h,w;
  cin>>h>>w;

  const int INF = 1e8;
  using P = pair<int, int>;
  set<P> sp;
  multiset<int> ans;

  auto push=[&](int j,int x){
    sp.emplace(j,x);
    ans.emplace(j+x);
  };

  push(0,INF);
  for(int j=1;j<=w;j++) push(j,-j);
  sp.emplace(w+1,-(w+1));

  auto erase=[&](P p){
    assert(sp.count(p));
    sp.erase(p);
    ans.erase(ans.find(p.first+p.second));
  };

  for(int i=0;i<h;i++){
    int a,b;
    cin>>a>>b;
    int v=INF;

    // erase points in [a, b]
    while(1){
      auto it=sp.lower_bound(P(a,-INF));
      if(b<it->first) break;
      v=it->second;
      erase(*it);
    }

    // add if last range exceed [a, b]
    if(v!=INF){
      auto it=sp.lower_bound(P(a,-INF));
      if(b+1<it->first) push(b+1,v);
    }

    if(ans.size()==1) cout<<-1<<newl;
    else cout<<i+1+*ans.begin()<<newl;
  }
  return 0;
}