#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
//consider the case where we only need k smallest from two different set of points
//if can do this in some reasonable time then apply zs bit trick
//+log factor
//need <= log^3 maybe?
//ok two set need log^2(?)
#define all(a) a.begin(),a.end()
#define pb push_back
const int MAXN = 2000000;
int ndcnt = 0;
int t[MAXN],ch[2][MAXN];
vector<ll>glo;
int main()
{
int n,k;
cin>>n>>k;
vector<array<int,2>>p(n);
vector<int>cy(n),cxy(n);
for(int i=0;i<n;i++)cin>>p[i][0]>>p[i][1];
sort(all(p));
vector<int>crd;
for(int i=0;i<n;i++)crd.push_back(p[i][1]);
sort(all(crd));
crd.resize(unique(all(crd)) - crd.begin());
for(int i=0;i<n;i++)cy[i] = lower_bound(all(crd),p[i][1]) - crd.begin();
for(int i=0;i<n;i++){
for(int j=0;j<i;j++){
ll c = abs(p[i][0]-p[j][0]);
c+=abs(p[i][1]-p[j][1]);
glo.pb(c);
}
}
sort(all(glo));
for(int i=0;i<k;i++)cout<<glo[i]<<'\n';
}
