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;
typedef long long ll;
const int N=1e6+10;
const int mod=1e9+7;
const int inf=0x3f3f3f3f;
ll n,cnt=0;
multiset<int> s;//multiset，允许重复，升序排列
map<int,int>mp;
int main() {
//    freopen("../in.in", "r", stdin);
//    freopen("../out.out","w",stdout);
    ios::sync_with_stdio(false);
    ll t, i, j = 0;
    ll m, x, y,k;
    cin>>n>>m;
    for(i=1;i<=m;i++) mp[i]=i,s.insert(0);//初始化
    for(i=1;i<=n;i++){
        int l,r;
        cin>>l>>r;
        auto it=mp.lower_bound(l);//从l开始
        int mx=0;
        while(it->first<=r&&it!=mp.end()){//循环到l-r区间结束
            mx=max(mx,it->second);//取离r最近的点
            s.erase(s.find(it->first-it->second));//将点从map和set中删去
            mp.erase(it++);
        }
        if(mx!=0&&r<m&&mp[r+1]<mx){//前两项防止越界，当map中不存在r+1时
            mp[r+1]=mx;//将r+1加入map和set
            s.insert(r+1-mx);
        }
        if(s.size()) cout<<*(s.begin())+i<<endl;//若set有点，直接取第一个+跨层步数i就是解
        else cout<<-1<<endl;
    }
    return 0;
}