/*
MLE

水の状態が同じでも4の倍数のインデックスが異なるなら別物と区別

*/
#include<iostream>
#include<set>
#include<algorithm>
#include<vector>
#include<string>
#include<set>
#include<map>
#include<numeric>
#include<queue>
#include<cmath>
#include<tuple>
using namespace std;
typedef long long ll;
const ll INF=1LL<<60;
typedef pair<ll,ll> P;
typedef pair<int,P> PP;
const ll MOD=1e9+7;

typedef tuple<int,int,int,int> TP;
//typedef vector<int> TP;

typedef pair<vector<int>,ll> VECP;




vector<int> V(4);

/*
TP func(int t1,int t2,int t3,int t4,ll idx){

    TP res=make_tuple(t1,t2,t3,t4);
    if(idx==0){
        int u1=t1-min(V2-t2,t1);
        int u2=t2+min(V2-t2,t1);
        res=make_tuple(u1,u2,t3,t4);
    }
    else if(idx==1){
        int u2=t2-min(V3-t3,t2);
        int u3=t3+min(V3-t3,t2);
        res=make_tuple(t1,u2,u3,t4);
    }else if(idx==2){
        int u3=t3-min(V4-t4,t3);
        int u4=t4+min(V3-t3,t3);
        res=make_tuple(t1,t2,u3,t4);
    }else if(idx==3){
        int u4=t3-min(V1-t1,t3);
        int u1=t1+min(V1-t1,t3);
        res=make_tuple(u1,t2,t3,u4);

    }

    return res;
}
*/

VECP func(const VECP& t){

    ll idx=t.second;
    vector<int> res(4);
    for(int i=0;i<4;i++){
        res[i]=t.first[i];
    }

    int w= min(V[(idx+1)%4] - t.first[(idx+1)%4],t.first[idx]);
    res[idx]=t.first[idx] - w;
    res[(idx+1)%4]=t.first[(idx+1)%4] + w;

    return make_pair(res,(idx+1)%4);
}


bool operator<(const VECP& lhs,const VECP& rhs){

    if(lhs.first!=rhs.first){
        return lhs.first<rhs.first;
    }else{
        return lhs.second<rhs.second;
    }
}

int main(){
    ll N;

    for(int i=0;i<4;i++){
        cin>>V[i];
    }
    cin>>N;

    vector<VECP> history;
    map<VECP,int> mp;



    VECP now=make_pair(vector<int>{V[0],0,0,0},0LL);

    int sz=0;

    int offset=-1;


    while(1){
        /*
        cout<<"sz="<<sz<<endl;
        for(int v:now){
            cout<<v<<' ';
        }
        cout<<endl;
        */
       if(sz==N) break;


       
        if(mp.count(now)){

            offset=mp[now];
            
            ll cyclelen=sz-offset;

            now=history[((N-mp[now])%cyclelen+mp[now])];
           

            break;

        }else{
            mp[now]=sz;

            history.push_back(now);

            auto nx=func(now);
            sz++;
            now=nx;
        }
        

     
    }

    
    for(int v:now.first){
        cout<<v<<' ';
    }
    cout<<endl;
    return 0;
        


}