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;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int H, W;
  cin >> H >> W;
  vector<int> A(H), B(H);
  for (int i = 0; i < H; i++) {
    cin >> A[i] >> B[i];
    A[i]--;
  }
  map<int, int> M;
  multiset<int> res;
  for (int i = 0; i < W; i++) {
    M[i] = i;
    res.insert(0);
  }
  for (int i = 0; i < H; i++) {
    auto it = M.lower_bound(A[i]);
    int s = -1;
    while (it != M.end() && it->first <= B[i]) {
      int x = it->first - it->second;
      s = max(s, it->second);
      res.erase(res.find(x));
      M.erase(it++);
    }
    if (s != -1 && B[i] < W) {
      M[B[i]] = s;
      res.insert(B[i] - s);
    }
    if (!res.empty()) {
      cout << *res.begin() + i + 1 << '\n';
    } else {
      cout << -1 << '\n';
    }
  }
  return 0;
}