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;
#define rep(i,n) for(int i = 0; i < (int)n; i++)
using ll = long long;
const ll INF = 1e9;
using P = pair<int,int>;

int main(){
  int h, w;
  cin >> h >> w;
  int a[h], b[h];
  rep(i,h) {
    cin >> a[i] >> b[i];
    a[i]--; b[i]--;
  }
  set<P> st;
  multiset<int> ms;
  rep(i,w) st.insert(P(i,i));
  rep(i,w) ms.insert(0);
  ms.insert(INF);
  rep(i,h) {
    vector<P> v;
    auto itr = st.lower_bound(P(a[i],0));
    while(itr != st.end() and (*itr).first <= b[i]) {
      v.push_back(*itr);
      itr = st.erase(itr);
    }
    for(auto p : v) ms.erase(ms.find(p.first-p.second));
    if((int)v.size()) {
      P p = v.back();
      if(b[i] < w-1) {
        st.insert(P(b[i]+1,p.second));
        ms.insert(b[i]+1-p.second);
      }
    }
    int ans = *ms.begin();
    cout << (ans == INF ? -1 : ans+i+1) << endl;
  }
  return 0;
}