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>
#define ll long long
#define db double
using namespace std;
const int INF=0x3f3f3f3f;
int h,w;
multiset<int> s;
map<int,int> mp;
int main()
{
    scanf("%d%d",&h,&w);
    for(int i=0;i<w;i++)
        s.insert(0),mp[i]=i;
    for(int i=0,x,y;i<h;i++)
    {
        scanf("%d%d",&x,&y);
        auto tmp=mp.lower_bound(x-1);
        int m=-1;
        while(tmp!=mp.end() && tmp->first<=y)
        {
            m=max(m,tmp->second);
            int p=(tmp->first)-(tmp->second);
            s.erase(s.find(p));
            mp.erase(tmp++);
        }
        if(m!=-1 && y<w)
        {
            s.insert(y-m);
            mp[y]=m;
        }
        printf("%d\n",s.empty()?-1:*s.begin()+i+1);
    }
    return 0;
}