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 (int i = 0; i < (n); i++)
#define sz(x) int(x.size())
using namespace std;
typedef long long ll;
const long long  mod = 1000000007;



int main(){
  int h,w;
  cin >> h >> w;

  map<int,int> s;
  multiset<int> v;

  rep(i,w) s[i] = i;
  rep(i,w) v.insert(0);

  rep(i,h){
    int r,l;
    cin >> l >> r;
    l--;

    auto it = s.lower_bound(l);
    int mr = -1;
    while(it != s.end() && it->first <=r){
      mr = max(mr, it->second);
      int x = it->first - it->second;
      v.erase(v.find(x));
      s.erase(it++);
      if(r > w-1) mr = -1;
    }
    if(mr != -1) {
      v.insert(r-mr);
      s[r] = mr;
    }
    int ans = -1;
    if(v.size()>0) {
      ans = *v.begin() + i+1;
    }
    cout << ans << endl;
  }
}