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 <algorithm>
#include <cmath>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <tuple>
#include <vector>
using namespace std;
#define rep(i, j, n) for (int i = j; i < (int)n; ++i)

int main() {
  int h, w;
  cin >> h >> w;

  multiset<int> st;
  map<int, int> mp;
  rep(i, 0, w) {
    mp[i] = i;
    st.insert(0);
  }

  int a, b;
  rep(i, 0, h) {
    cin >> a >> b;
    --a;
    int most_right = -1;
    for (auto itr = mp.lower_bound(a); itr != mp.upper_bound(b);) {
      most_right = max(most_right, itr->second);
      int d = itr->first - itr->second;
      st.erase(st.find(d));
      itr = mp.erase(itr);
    }
    if (most_right != -1 && b < w) {
      st.insert(b - most_right);
      mp[b] = most_right;
    }
    if (st.size() > 0)
      cout << *st.begin() + i + 1 << '\n';
    else
      cout << -1 << '\n';
  }
  return 0;
}