There are 2 special dices on the table. On each face of the dice, a distinct number was written. Consider a₁.a₂,a₃,a₄,a₅,a₆ to be numbers written on top face, bottom face, left face, right face, front face and back face of dice A. Similarly, consider b₁.b₂,b₃,b₄,b₅,b₆ to be numbers on specific faces of dice B. It’s guaranteed that all numbers written on dices are integers no smaller than 1 and no more than 6 while a_i ≠ a_j and b_i ≠ b_j for all i ≠ j. Specially, sum of numbers on opposite faces may not be 7.

At the beginning, the two dices may face different(which means there exist some i, a_i ≠ b_i). Ddy wants to make the two dices look the same from all directions(which means for all i, a_i = b_i) only by the following four rotation operations.(Please read the picture for more information)


Now Ddy wants to calculate the minimal steps that he has to take to achieve his goal.

There are multiple test cases. Please process till EOF. 

For each case, the first line consists of six integers a₁,a₂,a₃,a₄,a₅,a₆, representing the numbers on dice A. 

The second line consists of six integers b₁,b₂,b₃,b₄,b₅,b₆, representing the numbers on dice B.

For each test case, print a line with a number representing the answer. If there’s no way to make two dices exactly the same, output -1.

1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 5 6 4 3
1 2 3 4 5 6
1 4 2 5 3 6

0
3
-1

2014 ACM/ICPC Asia Regional Xi‘an Online

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代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<cmath>
#include<string>
#include<queue>
#define eps 1e-9
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;

const int maxn=999999+10;

struct cube
{
    int a1,a2,a3,a4,a5,a6;
    int key;
}st,en,next;

queue<cube>Q;
int ans;

bool vis[maxn];

int gao(cube c)
{
    int temp=0;
    temp+=c.a1*1;
    temp+=c.a2*10;
    temp+=c.a3*100;
    temp+=c.a4*1000;
    temp+=c.a5*10000;
    temp+=c.a6*100000;
    return temp;
}

cube solve(cube c,int k)
{
    cube temp;
    if(k==1)
    {
        temp.a1=c.a4;
        temp.a2=c.a3;
        temp.a3=c.a1;
        temp.a4=c.a2;
        temp.a5=c.a5;
        temp.a6=c.a6;
    }
    else if(k==2)
    {
        temp.a1=c.a3;
        temp.a2=c.a4;
        temp.a3=c.a2;
        temp.a4=c.a1;
        temp.a5=c.a5;
        temp.a6=c.a6;
    }
    else if(k==3)
    {
        temp.a1=c.a6;
        temp.a2=c.a5;
        temp.a3=c.a3;
        temp.a4=c.a4;
        temp.a5=c.a1;
        temp.a6=c.a2;
    }
    else
    {
        temp.a1=c.a5;
        temp.a2=c.a6;
        temp.a3=c.a3;
        temp.a4=c.a4;
        temp.a5=c.a2;
        temp.a6=c.a1;
    }
    return temp;
}

int bfs()
{
    ans=gao(en);
    memset(vis,false,sizeof(vis));
    while(!Q.empty())   Q.pop();
    st.key=0;
    Q.push(st);
    int temp=gao(st);
    vis[temp]=true;
    while(!Q.empty())
    {
        cube current=Q.front();
        Q.pop();
        if(ans==gao(current))
            return current.key;
        for(int i=1;i<=4;i++)
        {
           next=solve(current,i);
           next.key=current.key+1;
           int temp=gao(next);
           if(!vis[temp])
           {
               vis[temp]=true;
               Q.push(next);
           }
        }
    }
    return -1;
}


int main()
{
    while(~scanf("%d",&st.a1))
    {
        scanf("%d%d%d%d%d",&st.a2,&st.a3,&st.a4,&st.a5,&st.a6);
        scanf("%d%d%d%d%d%d",&en.a1,&en.a2,&en.a3,&en.a4,&en.a5,&en.a6);
        printf("%d\n",bfs());
    }
    return 0;
}