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 <iostream>
#include <vector>
#include <map>
#include <set>
using namespace std;
using ll = int64_t;

int main() {
  ll H,W;
  cin >> H >> W;
  ll INF = static_cast<ll>((H+1)*W);
  vector<ll> A(H, 0);
  vector<ll> B(H, 0);
  for(ll i=0; i<H; i++) {
    cin >> A[i] >> B[i];
  }
  map<ll,ll> M;
  multiset<ll> V;
  for(ll i=1; i<=W; i++) {
    M.insert(make_pair(i, 0));
    V.insert(0);
  }
  for(ll i=0; i<H; i++) {
    ll bp1 = INF;
    auto it = M.lower_bound(A[i]);
    while(true) {     
      if(it != M.end() && it->first<=B[i]) {
        bp1 = min(bp1, it->second + (B[i]+1-it->first));
        V.erase(V.find(it->second));
        M.erase(it++);
      } else {
        break;
      }
    }
    if(bp1 < INF && B[i]+1<=W) {
      if((M.count(B[i]+1)==0))
      	M[B[i]+1] = bp1,V.insert(bp1);
      else{
      	if(M[B[i]+1]>bp1)
          V.erase(V.find(M[B[i]+1])),M[B[i]+1]=bp1,V.insert(bp1);
      }
    }
    if(V.empty()) {		
      while(i<H)
		cout<<-1<<"\n",++i;
      return 0;
    } else {
      cout << ((*V.begin())+i+1) << endl;
    }
  }
}