/*

     题意：就是给定两个筛子，每个筛子上6个面，每个面的数字属于[1,6]， 且互不相同！

     问a筛子最少经过按照题目规定的要求转动，达到和b筛子上下左右前后的数字相同！

     思路：很直白的bfs，将每一种状态对应一个数字，保证这种状态不会重新加入队列中！

 */

 #include<iostream>

 #include<cstdio>

 #include<cstring>

 #include<algorithm>

 #include<queue>

 using namespace std;

 int a[], ss;

 int vis[];

 struct node{

     int k[];

     node(){}

     node(int a1, int a2, int a3, int a4, int a5, int a6){

         k[]=a1;

         k[]=a2;

         k[]=a3;

         k[]=a4;

         k[]=a5;

         k[]=a6;

     }

     int step;

 };

 queue<node>q; 

 int sum(node x){

     int s=;

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

        s= s* + x.k[i];

     return s;

 }

 bool bfs(){

     while(!q.empty()) q.pop();

     node cur(a[], a[], a[], a[], a[], a[]);

     cur.step=;

     q.push(cur);

     vis[sum(cur)]=;

     while(!q.empty()){

         cur = q.front();

         if(sum(cur)==ss){

             printf("%d\n", cur.step);

             return true;

         }

         q.pop();

         node *nt = new node(cur.k[], cur.k[], cur.k[], cur.k[], cur.k[], cur.k[]);

         int v = sum(*nt);

         if(!vis[v]){

             vis[v]=;

             nt->step = cur.step + ;

             q.push(*nt);

         }

         nt = new node(cur.k[], cur.k[], cur.k[], cur.k[], cur.k[], cur.k[]);

         v = sum(*nt);

         if(!vis[v]){

             vis[v]=;

             nt->step = cur.step + ;

             q.push(*nt);

         }

         nt = new node(cur.k[], cur.k[], cur.k[], cur.k[], cur.k[], cur.k[]);

         v = sum(*nt);

         if(!vis[v]){

             vis[v]=;

             nt->step = cur.step + ;

             q.push(*nt);

         }

         nt = new node(cur.k[], cur.k[], cur.k[], cur.k[], cur.k[], cur.k[]);

         v = sum(*nt);

         if(!vis[v]){

             vis[v]=;

             nt->step = cur.step + ;

             q.push(*nt);

         }

     }

     return false;

 }

 int main(){

     while(scanf("%d%d%d%d%d%d", &a[], &a[], &a[], &a[], &a[], &a[])!=EOF){

         ss=;

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

             int x;

             scanf("%d", &x);

             ss = ss *  + x;

         }

         memset(vis, , sizeof(vis));

         if(!bfs())

             printf("-1\n");

     }

     return ;

 }