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:
// -----------------------------------
// author   : MatsuTaku
// country  : Japan
// created  : 08/29/20 20:54:54
// -----------------------------------

#include <bits/stdc++.h>

using namespace std;
using ll = long long;

int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);

  int h,w; cin>>h>>w;

  map<int,int> etos;
  multiset<int> distset;
  for (int i = 0; i < w; i++) {
    etos[i] = i;
    distset.insert(0);
  }
  for (int i = 0; i < h; i++) {
    int a,b; cin>>a>>b; a--; b--;
    int max_s = -1;
    {
      auto bgn = etos.lower_bound(a);
      auto end = etos.upper_bound(b+1);
      for (auto it = bgn; it != end; ++it) {
        max_s = max(max_s, it->second);
        auto dist = it->first - it->second;
        assert(distset.count(dist)>0);
        distset.erase(distset.find(dist));
      }
      etos.erase(bgn, end);
    }
    if (max_s != -1 and b+1 < w) {
      auto min_dist = b+1 - max_s;
      distset.insert(min_dist);
      assert(etos.count(b+1) == 0);
      etos[b+1] = max_s;
    }
    if (!distset.empty()) {
      cout << i+1 + *distset.begin() << endl;
    } else {
      cout << -1 << endl;
    }
  }

  return 0;
}