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:
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
signed main() { 
  ios::sync_with_stdio(false); cin.tie(0);
  //f
  int h, w;
  cin >> h >> w;
  multiset<int> st;
  map<int, int> mp;
  for (int i = 0; i < w; i++) {
    st.insert(0);
    mp[i] = 0;
  }
  for (int i = 1; i <= h; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    int last = -1;
    int d = 1 << 30;
    auto it = mp.lower_bound(a);
    while (it != mp.end() && it->first < b) {
      assert(st.find(it->second) != st.end());
      st.erase(st.find(it->second));
      last = it->first;
      d = min(d, it->second + b - it->first);
      it = mp.erase(it);
    }
    if (last != -1) {
      if (b < w) {
        if (mp.count(b)) {
          st.erase(st.find(mp[b]));
          mp[b] = min(d, mp[b]);
          st.insert(mp[b]);
        } else {
          mp[b] = d;
          st.insert(d);
        }
      }
    }
    int ans = (!st.empty() ? *st.begin() + i : -1);
    cout << ans << '\n';
  }
  return 0;
}