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>
#define rep(i, n) for (decltype(+n) i = 0; i < (n); i++)
using namespace std;

using vi = vector<int>;
using pii = pair<int, int>;

template <class T>
void COUT(const T &x) { cout << x << endl; }
template <class T>
bool amax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}

signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);

  int H, W;
  cin >> H >> W;

  set<pii> rl;
  rep(i, W)rl.emplace(i, i);
  multiset<int> ds;
  rep(_, W)ds.emplace(0);

  rep(k, H) {
    int a, b;
    cin >> a >> b;
    a--;
    int l = -1;
    for(auto it = rl.lower_bound(pii(a, 0)); it != rl.cend() && it->first <= b; it = rl.erase(it)) {
      ds.erase(ds.find(it->first - it->second));
      amax(l, it->second);
    }
    if(l > -1 && b < W) {
      rl.emplace(b, l);
      ds.emplace(b - l);
    }
    COUT(ds.empty() ? -1 : *ds.cbegin() + k + 1);
  }
}