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 <iostream>
#include <string>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <math.h>
#include <cassert>
#define rep(i,n) for(int i = 0; i < n; ++i )
using namespace std;
using ll = long long;
using P = pair<int,int>;

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

  map<int,int> t;
  multiset<int> s;
  rep(i,w) t[i]=i, s.insert(0);
  rep(k,h) {
    int l,r;
    cin >> l >> r;
    int d = -1;
    auto i=t.lower_bound(--l);
    while(i!=t.end() && i->first<=r){
      d = max(d,i->second);
      s.erase(s.find(i->first - i->second));
      t.erase(i++);
    }
    if(0<=d && r<w){
      s.insert(r-d);
      t[r] = d;
    }
    int ans = s.empty() ? -1 : *s.begin() + k + 1;
    cout << ans << endl;
  }
}