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:
// #pragma GCC optimize(3,"Ofast","inline")
#include<bits/stdc++.h>
//#include<ctime>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int mod=1e9+7;
const int MAXN=1e6+7;
///////////////////////////////////////////////////////////////////////////////////////
map<int,int> q;
multiset<int> p;
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    for (int i=0;i<m;i++){
        q[i]=i;
        p.insert(0);
    }
    for (int i=1;i<=n;i++){
        //cout << "i=" << i << endl;
        int x,y;
        scanf("%d%d",&x,&y);
        auto f=q.lower_bound(--x);
        int ff=-1;
        while(f!=q.end()&&f->first<=y){
            ff=max(ff,f->second);
            p.erase(p.find(f->first-f->second));
            q.erase(f++);
        }
        if(ff!=-1&&y<m){
            p.insert(y-ff);
            q[y]=ff;
        }
        if(p.size()==0) printf("-1\n");
        else printf("%d\n",*p.begin()+i);
    }
    return 0 ;
}