inline void dp()//动态规划，强行枚举八种情况 
{
    f[0][0][0][0][0]=0;
    for (int i=0;i<n;++i)
        for (int j=0;j<=jla;++j)
            for (int k=0;k<=jlb;++k)
                for (int l=0;l<=jlc;++l)
                {
                    long long tmp=f[i][j][k][l][0];//枚举最后一位不进位的情况
                    f[i+1][j+1][k+1][l+1][1]=min(f[i+1][j+1][k+1][l+1][1],tmp+(1<<i+1));
                    f[i+1][j+1][k][l+1][0]=min(f[i+1][j+1][k][l+1][0],tmp+(1<<i));
                    f[i+1][j][k+1][l+1][0]=min(f[i+1][j][k+1][l+1][0],tmp+(1<<i));
                    f[i+1][j][k][l][0]=min(f[i+1][j][k][l][0],tmp);
                    tmp=f[i][j][k][l][1];//枚举最后一位进位的情况
                    f[i+1][j+1][k+1][l+1][1]=min(f[i+1][j+1][k+1][l+1][1],tmp+(1<<i+1));
                    f[i+1][j][k+1][l][1]=min(f[i+1][j][k+1][l][1],tmp+(1<<i));
                    f[i+1][j+1][k][l][1]=min(f[i+1][j+1][k][l][1],tmp+(1<<i));
                    f[i+1][j][k][l][0]=min(f[i+1][j][k][l][0],tmp);
                }
}

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
#define INF 0x7f7f7f7f7f7f7f
int T,a,b,c;
int n,jla,jlb,jlc;//n记录三个数的二进制数码长度的最大值，jla、jlb、jlc分别记录a、b、c的二进制数码 
long long f[33][33][33][33][2];
inline int read()//快读 
{
    char kr=0;
    char ls;
    for(;ls>‘9‘||ls<‘0‘;kr=ls,ls=getchar());
    int xs=0;
    for(;ls>=‘0‘&&ls<=‘9‘;ls=getchar())
    {
        xs=xs*10+ls-48;
    }
    if(kr==‘-‘) xs=0-xs;
    return xs;
}
inline int lowbit(int x)//求出x的二进制数码 
{
    int sum=0;
    for (;x;x>>=1)
        sum+=x&1;
    return sum;
}
inline void dp()//动态规划，强行枚举八种情况 
{
    f[0][0][0][0][0]=0;
    for (int i=0;i<n;++i)
        for (int j=0;j<=jla;++j)
            for (int k=0;k<=jlb;++k)
                for (int l=0;l<=jlc;++l)
                {
                    long long tmp=f[i][j][k][l][0];//枚举最后一位不进位的情况
                    f[i+1][j+1][k+1][l+1][1]=min(f[i+1][j+1][k+1][l+1][1],tmp+(1<<i+1));
                    f[i+1][j+1][k][l+1][0]=min(f[i+1][j+1][k][l+1][0],tmp+(1<<i));
                    f[i+1][j][k+1][l+1][0]=min(f[i+1][j][k+1][l+1][0],tmp+(1<<i));
                    f[i+1][j][k][l][0]=min(f[i+1][j][k][l][0],tmp);
                    tmp=f[i][j][k][l][1];//枚举最后一位进位的情况
                    f[i+1][j+1][k+1][l+1][1]=min(f[i+1][j+1][k+1][l+1][1],tmp+(1<<i+1));
                    f[i+1][j][k+1][l][1]=min(f[i+1][j][k+1][l][1],tmp+(1<<i));
                    f[i+1][j+1][k][l][1]=min(f[i+1][j+1][k][l][1],tmp+(1<<i));
                    f[i+1][j][k][l][0]=min(f[i+1][j][k][l][0],tmp);
                }
}
inline void clear()//为做DP初始化 
{
    memset(f,INF,sizeof(f)); 
    n=max((int)log2(a)+1,(int)log2(b)+1);
    n=max(n,(int)log2(c)+1);//求 n 
    jla=lowbit(a),jlb=lowbit(b),jlc=lowbit(c);
}
int main()
{
    a=read();b=read();c=read();
    clear();
    dp();
    if(f[n][jla][jlb][jlc][0]>=INF)//注意是“≥INF” 
    {
        printf("-1\n");
        return 0;
    }//如果无解就输出-1 
    printf("%lld\n",f[n][jla][jlb][jlc][0]);//输出最小值 
    return 0;
}