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Pattern lock security is generally used in Android handsets instead of a password. The pattern lock can be set by joining points on a 3 × 3 matrix in a chosen order. The points of the matrix are registered in a numbered order starting with 1 in the upper left corner and ending with 9 in the bottom right corner.

A valid pattern has the following properties:

• A pattern can be represented using the sequence of points which it's touching for the first time (in the same order of drawing the pattern). And we call those points as active points.For every two consecutive points A and B in the pattern representation, if the line segment connecting A and B passes through some other points, these points must be in the sequence also and comes before A and B, otherwise the pattern will be invalid.In the pattern representation we don't mention the same point more than once, even if the pattern will touch this point again through another valid segment, and each segment in the pattern must be going from a point to another point which the pattern didn't touch before and it might go through some points which already appeared in the pattern.

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  Now you are given n active points, you need to find the number of valid pattern locks formed from those active points.

  Input

  There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

  The first line contains an integer n (3 ≤ n ≤ 9), indicating the number of active points. The second line contains n distinct integers a₁, a₂, … a_n (1 ≤ a_i ≤ 9) which denotes the identifier of the active points.

  Output

  For each test case, print a line containing an integer m, indicating the number of valid pattern lock.

  In the next m lines, each contains n integers, indicating an valid pattern lock sequence. The m sequences should be listed in lexicographical order.

  Sample Input

  1
  3
  1 2 3
  

  Sample Output

  4
  1 2 3
  2 1 3
  2 3 1
  3 2 1
  

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  Author: LIN, Xi
  Source: The 15th Zhejiang University Programming Contest
  Submit Status
  

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  #include 
      
       
  #include 
       
         #include 
        
          #include 
         
           #include 
          
            #include 
           
             using namespace std; set
            
              st; vector
             
               vi; bool vis[20]; int n; bool go[30]; bool graph[5][5]; int XY[10][2]={{0,0},{1,1},{1,2},{1,3},{2,1},{2,2},{2,3},{3,1},{3,2},{3,3}}; int TU[4][4]= { 0,0,0,0, 0,1,2,3, 0,4,5,6, 0,7,8,9 }; const int dir_x[16]={-1,-1,-1,0,0,1,1,1,-1,-1,1,1,-2,-2,2,2}; const int dir_y[16]={-1,0,1,-1,1,-1,0,1,-2,2,-2,2,-1,1,-1,1}; bool isIn(int a,int b) { if((a<=0||a>3)||(b<=0||b>3)) return false; return true; } int as[10]; void output(int x) { int na=0; while(x) { as[na++]=x%10; x/=10; } for(int i=na-1;i>=0;i--) printf(%d%c,as[i],(i==0)?10:32); } void dfs(int x,int y,int u,int nb,int deep) { if(deep==n-1) { if(st.count(nb)==0) { st.insert(nb); vi.push_back(nb); } return ; } for(int i=0;i<16;i++) { for(int j=1;j<=2;j++) { if(j==1) { int nx=x+dir_x[i],ny=y+dir_y[i]; if(isIn(nx,ny)==false) continue; int v=TU[nx][ny]; if(go[v]==false) continue; if(vis[v]==true) continue; vis[v]=true; dfs(nx,ny,v,nb*10+v,deep+1); vis[v]=false; } else if(j==2) { int nx=x+dir_x[i]*2,ny=y+dir_y[i]*2; if(isIn(nx,ny)==false) continue; /// check mid int xx=x+dir_x[i],yy=y+dir_y[i]; int tv=TU[xx][yy],vv=TU[nx][ny]; if(vis[tv]==false) continue; if(go[tv]==false) continue; if(vis[vv]==true) continue; if(go[vv]==false) continue; vis[vv]=true; dfs(nx,ny,vv,nb*10+vv,deep+1); vis[vv]=false; } } } } int main() { int T_T; scanf(%d,&T_T); while(T_T--) { scanf(%d,&n); memset(graph,false,sizeof(graph)); memset(go,false,sizeof(go)); st.clear(); vi.clear(); for(int i=0;i
              
               
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