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 <cassert>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <limits.h>
#include <map>
#include <queue>
#include <set>
#include <string.h>
#include <vector>

using namespace std;
typedef long long ll;

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

  map<int, int> dests;
  multiset<int> lengths;

  for (int x = 1; x <= W; ++x) {
    dests[x] = x;
    lengths.insert(0);
  }

  int a, b;
  for (int k = 1; k <= H; ++k) {
    cin >> a >> b;

    auto it = dests.lower_bound(a);
    int max_f = -1;

    while (it != dests.end() && it->first <= b + 1) {
      int dest = it->first;
      int from = it->second;

      max_f = max(max_f, from);
      int len = dest - from;
      // fprintf(stderr, "len: %d\n", len);
      lengths.erase(lengths.find(len));

      dests.erase(it++);
    }

    if (max_f != -1 && b < W) {
      lengths.insert(b - max_f + 1);
      dests[b + 1] = max_f;
    }

    /*
    for (auto it : dests) {
      fprintf(stderr, "%d -> %d\n", it.second, it.first);
    }
    */

    if (lengths.empty()) {
      cout << -1 << endl;
    } else {
      cout << *lengths.begin() + k << endl;
    }
  }

  return 0;
}