1 /*
  2     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！
  3     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！
  4
  5     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！
  6 */
  7 #include<iostream>
  8 #include<cstdio>
  9 #include<cstring>
 10 #include<algorithm>
 11 #include<queue>
 12 using namespace std;
 13
 14 int a[7], ss;
 15 int vis[654330];
 16
 17 struct node{
 18     int k[7];
 19     node(){}
 20     node(int a1, int a2, int a3, int a4, int a5, int a6){
 21         k[1]=a1;
 22         k[2]=a2;
 23         k[3]=a3;
 24         k[4]=a4;
 25         k[5]=a5;
 26         k[6]=a6;
 27     }
 28     int step;
 29 };
 30
 31 queue<node>q;
 32
 33
 34 int sum(node x){
 35     int s=0;
 36     for(int i=1; i<=6; ++i)
 37        s= s*10 + x.k[i];
 38     return s;
 39 }
 40
 41 bool bfs(){
 42     while(!q.empty()) q.pop();
 43     node cur(a[1], a[2], a[3], a[4], a[5], a[6]);
 44     cur.step=0;
 45     q.push(cur);
 46     vis[sum(cur)]=1;
 47     while(!q.empty()){
 48         cur = q.front();
 49         if(sum(cur)==ss){
 50             printf("%d\n", cur.step);
 51             return true;
 52         }
 53         q.pop();
 54         node *nt = new node(cur.k[5], cur.k[6], cur.k[3], cur.k[4], cur.k[2], cur.k[1]);
 55         int v = sum(*nt);
 56         if(!vis[v]){
 57             vis[v]=1;
 58             nt->step = cur.step + 1;
 59             q.push(*nt);
 60         }
 61
 62         nt = new node(cur.k[6], cur.k[5], cur.k[3], cur.k[4], cur.k[1], cur.k[2]);
 63         v = sum(*nt);
 64         if(!vis[v]){
 65             vis[v]=1;
 66             nt->step = cur.step + 1;
 67             q.push(*nt);
 68         }
 69
 70         nt = new node(cur.k[3], cur.k[4], cur.k[2], cur.k[1], cur.k[5], cur.k[6]);
 71         v = sum(*nt);
 72         if(!vis[v]){
 73             vis[v]=1;
 74             nt->step = cur.step + 1;
 75             q.push(*nt);
 76         }
 77
 78         nt = new node(cur.k[4], cur.k[3], cur.k[1], cur.k[2], cur.k[5], cur.k[6]);
 79         v = sum(*nt);
 80         if(!vis[v]){
 81             vis[v]=1;
 82             nt->step = cur.step + 1;
 83             q.push(*nt);
 84         }
 85     }
 86     return false;
 87 }
 88
 89 int main(){
 90     while(scanf("%d%d%d%d%d%d", &a[1], &a[2], &a[3], &a[4], &a[5], &a[6])!=EOF){
 91         ss=0;
 92         for(int i=1; i<=6; ++i){
 93             int x;
 94             scanf("%d", &x);
 95             ss = ss * 10 + x;
 96         }
 97         memset(vis, 0, sizeof(vis));
 98         if(!bfs())
 99             printf("-1\n");
100     }
101     return 0;
102 }