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 <map>
#include <set>
using namespace std;

int main() {
  int H, W;
  cin >> H >> W;
  map<int, int> m;
  multiset<int> s;
  for (int w = 1; w <= W; ++w) {
    m[w] = w;
    s.insert(0);
  }

  int a, b;
  for (int h = 1; h <= H; ++h) {
    cin >> a >> b;
    auto it = m.lower_bound(a);
    int mx = -1;
    while (it != m.end() && it->first <= b + 1) {
      mx = max(mx, it->second);
      int d = it->first - it->second;
      s.erase(s.find(d));
      m.erase(it++);
    }

    if (mx != -1 && b + 1 <= W) {
      s.insert(b+1 - mx);
      m[b+1] = mx;
    }
    if (s.empty()) {
      cout << -1 << '\n';
    } else {
      cout << *s.begin() + h << '\n';
    }
  }
  return 0;
}