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 rep(i,n) for (long long i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<ll,ll>;
using vec = vector<ll>;
using vecp = vector<P>;
using mat = vector<vec>;
using matp = vector<vecp>;
const ll MOD = 1e9+7;
const ll INF = 1e18;
#define all(v) v.begin(), v.end()

int main(){
  ll H,W;
  cin >> H >> W;
  vec A(H),B(H);
  rep(i,H){
    cin >> A.at(i) >> B.at(i);
  }
  set<P> C,D;
  rep(i,W){
    C.insert(P(i+1,0));
    D.insert(P(0,i+1));
  }
  rep(i,H){
    if(C.empty()){
      cout << -1 << endl;
    }else{
      P x=P(0,0);
      while(C.lower_bound(P(A.at(i),0))!=C.end()&&*C.lower_bound(P(A.at(i),0))<P(B.at(i)+1,0)){
        P a=*C.lower_bound(P(A.at(i),0));
        C.erase(a);
        D.erase(P(a.second,a.first));
        if(x==P(0,0)){
          x=a;
        }else{
          if(a.first-x.first>a.second-x.second){
            x=a;
          }
        }
        if(C.lower_bound(P(A.at(i),0))==C.end()||*C.lower_bound(P(A.at(i),0))>=P(B.at(i)+1,0)){
          if(B.at(i)!=W){
            C.insert(P(B.at(i)+1,x.second+B.at(i)+1-x.first));
            D.insert(P(x.second+B.at(i)+1-x.first,B.at(i)+1));
          }
        }
      }
      if(C.empty()){
        cout << -1 << endl;
      }else{
        P a=*begin(D);
        cout << a.first+i+1 << endl;
      }
    }
  }
}