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 int64_t ll;
#define rep(i,n) for(ll i=0;i<(ll)(n);i++)
ll h,w,a,b;
map<ll,ll> c;
multiset<ll> d;

int main(){
  cin>>h>>w;
  rep(i,w)c[i]=i;
  rep(i,w)d.insert(0);
  rep(i,h){
    cin>>a>>b;
    a--;
    auto e=c.lower_bound(a);
    ll m=-1;
    while(e!=c.end() && e->first<=b){
      m=max(m,e->second);
      ll x=(e->first)-(e->second);
      d.erase(d.find(x));
      c.erase(e++);
    }
    if(m!=-1 && b<w){
      d.insert(b-m);
      c[b]=m;
    }
    cout<<(d.empty() ? -1 : *d.begin()+i+1)<<endl;
  }
}