/*

TASK: milk3

LANG: C++

SOLVE: 倒水，dfs，枚举每一种倒法，ca[i][j]记录a和c桶的状态，因为总体积不变，故b的状态不需要记录。

 */

#include<cstdio>

#include<iostream>

#include<cstring>

#include<algorithm>

using namespace std;

int a,b,c;

bool ca[][];

void dfs(int na,int nb,int nc){

//    printf("%d %d %d\n",na,nb,nc);

    if(ca[na][nc])return;

    ca[na][nc]=;

    dfs(max(na+nb-b,),min(b,na+nb),nc);//a->b

    dfs(min(a,na+nb),max(na+nb-a,),nc);//b->a

    dfs(max(na+nc-c,),nb,min(c,na+nc));//a->c

    dfs(min(a,na+nc),nb,max(na+nc-a,));//c->a

    dfs(na,max(nc+nb-c,),min(c,nc+nb));//b->c

    dfs(na,min(nc+nb,b),max(nc+nb-b,));//c->b

}

int main(){

    freopen("milk3.in","r",stdin);

    freopen("milk3.out","w",stdout);

    scanf("%d%d%d",&a,&b,&c);

    dfs(,,c);

    int ok=;

    for(int i=;i<=c;i++)

        if(ca[][i]){

            if(ok) printf(" ");

            printf("%d",i);

            ok=;

        }

    puts("");

    return ;

}