#include <bits/stdc++.h> using namespace std; // #include <ext/pb_ds/assoc_container.hpp> // #include <ext/pb_ds/tree_policy.hpp> // using namespace __gnu_pbds; typedef long long ll; typedef long double ld; typedef pair<int, int> pii ; typedef pair<ll, ll> pll ; typedef vector<pii> vii ; typedef vector<int> veci ; typedef vector<pll> vll ; typedef vector<ll> vecll; // find_by_order order_of_key //#pragma GCC optimize("O3,unroll-loops") //#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt") #define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update> #define F first #define S second #define pb push_back #define endl '\n' #define Mp make_pair #define all(x) x.begin(), x.end() #define debug(x) cerr << #x << " = " << x << endl #define set_dec(x) cout << fixed << setprecision(x); #define fast_io ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); #define file_io freopen("in.txt" , "r" , stdin) ; freopen("out.txt" , "w" , stdout); #define lb lower_bound #define ub upper_bound #define for1(n) for(int i=1;i<=n;i++) #define for0(n) for(int i=0;i<n;i++) #define forn(n) for(int i=n;i>0;i--) #define pq priority_queue <pii, vector<pii>, greater<pii>> const ll mod = 1e9+7 ;// 998244353 ;// 1e9+9; ll inf=1e18; const int N=1e6+100,L=21,bs=701,NN=1e6; int B[N],C[N],D[N],E[N],n,m,k,q,dp[N],pre[N],dist[N],vis[N]; vector<int>g[N]; vector<ll>ans; pii A[N]; bool check(ll t){ set<pii>C; int r=1; for1(n){ while(A[r].F-A[i].F<=t && r<=n){ auto it=C.lb({A[r].S-t,-mod*2}); while(1){ if(it==C.end())break; ll dis=max(max((ll((*it).F)-ll(A[r].S)),-(ll((*it).F)-ll(A[r].S))),max((ll((*it).S)-ll(A[r].F)),-(ll((*it).S)-ll(A[r].F)))); if(dis>t)break; ans.pb(dis); if(ans.size()==k)return 1; it++; } C.insert({A[r].S,A[r].F}); r++; } C.erase({A[i].S,A[i].F}); } return 0; } int main(){ fast_io cin>>n>>k; for1(n){ int x,y;cin>>x>>y; A[i]={x+y,x-y}; } sort(A+1,A+n+1); ll l=0,r=r=mod*4; while(r-l!=1){ ll mid=(l+r)>>1; if(check(mid)==0)l=mid; else r=mid; ans.clear(); } check(l); sort(all(ans)); for(int x:ans)cout<<x<<" ";cout<<endl; for1(k-ans.size())cout<<l+1<<endl; }