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 a[200020],b[200020];

struct edge{
    int x,y;
    bool operator<(const edge& t)const{
        return y<t.y;
    }
};

set<edge> s;
multiset<int> ms;

int main() {
    int h,w;
    cin >> h >> w;
    for(int i=1;i<=h;i++)cin >> a[i] >> b[i];
    for(int i=1;i<=w;i++){
        s.insert({i,i});
        ms.insert(0);
    }
    for(int i=1;i<=h;i++){
        int ma=0;
        auto u=s.lower_bound({0,a[i]});
        while(u!=s.end()&&u->y<=b[i]+1){
            auto v=ms.find(u->y-u->x);
            ms.erase(v);
            ma=max(ma,u->x);
            u=s.erase(u);
        }
        if(ma>0&&b[i]<w){
            s.insert({ma,b[i]+1});
            ms.insert(b[i]+1-ma);
        }
        if(s.empty()){
            cout << "-1" << endl;
        } else {
            cout << *(ms.begin())+i << endl;
        }
    }
    return 0;
}