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;

using vi = vector<int>;
using ii = pair<int, int>;

int main() {
  cin.sync_with_stdio(false);
  cin.tie(NULL);

  int w, h;

  cin >> h >> w;
  set<ii> x, y;

  for (int i = 1; i <= w; i++) {
    y.insert(make_pair(0, i));
    x.insert(make_pair(i, i));
  }

  int a, b;
  for (int i = 1; i <= h; i++) {
    cin >> a >> b;
    ii st = make_pair(a, -1);
    ii end = make_pair(b, INT_MAX);
    for (auto it = x.lower_bound(st); it != x.upper_bound(end); it = x.lower_bound(st)) {
      int j = it->second;
      y.erase(make_pair(it->first - it->second, it->second));
      x.erase(it);
      
      if (x.lower_bound(st) == x.upper_bound(end)) {
        x.insert(make_pair(b == w ? INT_MAX : b + 1, j));
        y.insert(make_pair(b == w ? INT_MAX : b + 1 - j, j));
      }
    }

    int ans = y.begin()->first;
    cout << (ans == INT_MAX ? -1 : ans + i) << endl;
  }

  return 0;
}