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() {
  int H, W;
  cin >> H >> W;

  map<int, int> MA;
  multiset<int> MS;
  for (int i = 0; i < W; i++) {
    MA[i] = i;
    MS.insert(0);
  }

  for (int i = 0; i < H; i++) {
    int A, B;
    cin >> A >> B, A--;
    auto it = MA.lower_bound(A);
    int tmp = -1;
    while (it != MA.end() && it -> first <= B) {
      tmp = max(tmp, it -> second);
      MS.erase(MS.find(it -> first - it -> second));
      MA.erase(it++);
    }
    if (tmp != -1 && B < W) {
      MA[B] = tmp;
      MS.insert(B - tmp);
    }
    if (MS.empty()) cout << -1 << "\n";
    else cout << *MS.begin() + i + 1 << "\n";
  }
}