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(0);
  int n, m;
  cin >> n >> m;
  vector<int> dist(m);
  set<int> st;
  multiset<int> all;
  for (int i = 0; i < m; i++) {
    st.insert(i);
    all.insert(0);
  }
  const int inf = (int) 2e9;
  for (int i = 0; i < n; i++) {
    int l, r;
    cin >> l >> r;
    --l;
    int Min = inf;
    auto it = st.lower_bound(l);
    while (it != st.end() && *it <= r) {
      int x = *it;
      Min = min(Min, dist[x] + r - x);
      it = st.erase(it);
      all.erase(all.find(dist[x]));
      dist[x] = inf;
    }
    if (r != m && Min != inf) {
      dist[r] = min(dist[r], Min);
      st.insert(r);
      all.insert(dist[r]);
    }
    if (all.empty()) {
      cout << -1 << '\n';
    } else {
      cout << *all.begin() + i + 1 << '\n';
    }
  }
  return 0;
}