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;
template <typename T> bool chmax(T &u, const T z) { if (u < z) {u = z; return true;} else return false; }
template <typename T> bool chmin(T &u, const T z) { if (u > z) {u = z; return true;} else return false; }
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
//解説写し
int main(){
  int h,w;
  cin>>h>>w;
  map<int,int>s;
  multiset<int>vals;
  rep(i,w)s[i]=i;
  rep(i,w)vals.insert(0);
  rep(i,h){
    int l,r;
    cin>>l>>r;
    --l;
    auto it=s.lower_bound(l);
    int mr=-1;
    while(it!=s.end() && it ->first <=r){
      chmax(mr,it->second);
      int x=it->first - it->second;
      vals.erase(vals.find(x));
      s.erase(it++);
    }
    if(mr!=-1 && r<w){
      vals.insert(r-mr);
      s[r]=mr;
    }
    int ans=-1;
    if(vals.size()>0)ans=*vals.begin()+i+1;
    printf("%d\n",ans);
  }
  return 0;
}