1 #include<iostream>
 2 #include<string>
 3 #include<cstring>
 4 #include<cstdio>
 5 #include<fstream>
 6 #include<iosfwd>
 7 #include<sstream>
 8 #include<fstream>
 9 #include<cwchar>
10 #include<iomanip>
11 #include<ostream>
12 #include<vector>
13 #include<cstdlib>
14 #include<queue>
15 #include<set>
16 #include<ctime>
17 #include<algorithm>
18 #include<complex>
19 #include<cmath>
20 #include<valarray>
21 #include<bitset>
22 #include<iterator>
23 #define ll long long
24 using namespace std;
25 const double clf=1e-8;
26 //const double e=2.718281828;
27 const double PI=3.141592653589793;
28 const int MMAX=2147483647;
29 //priority_queue<int>p;
30 //priority_queue<int,vector<int>,greater<int> >pq;]
31 int times,a,b,c,k;
32 struct node
33 {
34     int x,y,z,step;
35 };
36 int vis[51][51][51];
37 int dir[6][3]={{0,0,1},{0,0,-1},{0,-1,0},{0,1,0},{1,0,0},{-1,0,0}};//六个自由度的矢量
38 int bfs(int x,int y,int z,int x1,int x2,int x3)
39 {
40     int i;
41     queue<node> q;
42     q.push(node{x,y,z,0});
43     vis[x][y][z]=1;
44     while(!q.empty())
45     {
46         node t=q.front();//读最开始的那个元素
47         q.pop();//取出来
48         if(t.step>times)
49             return -1;//到不了返回-1
50         if(t.x==a-1&&t.y==b-1&&t.z==c-1&&t.step<=times)
51             return t.step;//到了直接返回步数
52         for(i=0;i<6;i++)//6个维度
53         {
54             int dx=t.x+dir[i][0];//x方向动
55             int dy=t.y+dir[i][1];//y方向动
56             int dz=t.z+dir[i][2];//z方向动
57             if(dx>=0&&dy>=0&&dz>=0&&dx<=x1&&dy<=x2&&dz<=x3&&!vis[dx][dy][dz])
58             {
59                 q.push(node{dx,dy,dz,t.step+1});//搜索到的放进去
60                 vis[dx][dy][dz]=1;
61             }
62         }
63     }
64     return -1;
65 }
66 int main()
67 {
68     scanf("%d",&k);
69     while(k--)
70     {
71         scanf("%d%d%d%d",&a,&b,&c,&times);
72         for(int i=0;i<a;i++)
73         {
74             for(int j=0;j<b;j++)
75             {
76                 for(int k=0;k<c;k++)
77                 {
78                     scanf("%d",&vis[i][j][k]);
79                 }
80             }
81         }
82         if(a==1&&b==1&&c==1)
83             printf("0\n");
84         else
85         {
86             int ans=bfs(0,0,0,a-1,b-1,c-1);//0,0,0开始a-1,b-1,c-1结尾
87                 printf("%d\n",ans);
88         }
89     }
90     return 0;
91 }