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 ll = long long;
using ull = unsigned long long;
using namespace std;
constexpr int inf = 1<<30;
constexpr int mo = 1e9+7;
constexpr ll infl = 1ll<<60;


int main(){
    cin.tie(nullptr);
    std::ios::sync_with_stdio(false);
    
    int h,w;
    cin>>h>>w;
    map<int,int> m;
    multiset<int> s;
    for(int i=0;i<w;++i)m[i]=i,s.insert(0);
    for(int i=0;i<h;++i){
        int a,b;
        cin>>a>>b;
        a--;
        auto it = m.lower_bound(a);
        int ma = -1;
        while(it != m.end() and it->first <= b){
            ma = max(ma,it->second);
            s.erase(s.find(it->first - it->second));
            m.erase(it++);
        }
        if(ma != -1 and b < w){s.insert(b-ma);m[b] = ma;}
        
        int ret =-1;
        if(s.size() > 0)ret = *s.begin() + (i+1);
        cout << ret << endl;
    }
    return 0;
    
}