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(){
    int h,w;
    cin>>h>>w;
    set<pair<int,int>>s;
    multiset<int>ms;
    for(int i=0;i<w;i++){
        s.insert({i,i});
        ms.insert(0);
    }
    vector<int>ans;
    for(int i=1;i<=h;i++){
        int a,b;
        cin>>a>>b;
        int max_start=-1;
        for(auto j=s.lower_bound({a-1,0});j!=s.lower_bound({b,0});j=s.erase(j)){
            max_start=max(max_start,(*j).second);
            ms.erase(ms.find((*j).first-(*j).second));
        }
        if(b<w && max_start!=-1){
            s.insert({b,max_start});
            ms.insert(b-max_start);
        }
        ans.push_back(ms.size() ? *ms.begin()+i : -1);
    }
    for(int i:ans)cout<<i<<endl;
    return 0;
}