1 #include <string.h>
 2 #include <stdio.h>
 3 #define MOD 1000000007
 4 #define MAXN 71
 5 typedef __int64 LL;
 6 int cas,n;
 7 int dmat[MAXN][MAXN],id[256],ids;
 8 struct matrix{
 9     int mz[MAXN][MAXN]; int n,tmp_cal[35];
10     #define FOR(i) for(int i=1;i<=n;i++)
11     //初始化矩阵,空矩阵,单位矩阵和dmat矩阵
12     void init(int nn,int type){
13         memset(tmp_cal,0,sizeof tmp_cal);n=nn;
14         if(type==0)FOR(i)FOR(j)mz[i][j]=0;
15         else if(type==1)FOR(i)FOR(j)mz[i][j]=(i==j)?1:0;
16         else FOR(i)FOR(j)mz[i][j]=dmat[i][j];
17     }
18     matrix operator *(const matrix& b){
19         matrix ans;ans.init(n,0);
20         FOR(i)FOR(j)if(mz[i][j])FOR(k)
21             ans.mz[i][k]=(ans.mz[i][k]+(LL)mz[i][j]*b.mz[j][k])%MOD;
22         return ans;
23     }
24     matrix binMat(int x);
25 }tmp[35],m;
26 matrix matrix::binMat(int x){
27     if(!tmp_cal[0]){tmp_cal[0]=1,tmp[0].init(n,2);}
28     matrix ans;ans.init(this->n,1);
29     for(int i=0;x;x>>=1,i++){
30         if(x&1)ans=ans*tmp[i];
31         if(!tmp_cal[i+1])tmp_cal[i+1]=1,tmp[i+1]=tmp[i]*tmp[i];
32     }
33     return ans;
34 }
35 //判断是否可接的条件,或的结果为1<<8-1,与的结果中没有连续的奇数个1
36 int cango(int p1,int p2){
37     if((p1|p2)!=255)return 0;
38     p1=p1&p2;if(p1==255)return 1;
39     while(p1&1)p1=1<<7|(p1>>1);
40     for(int i=0,d=0;i<9;i++)
41         if(i<8&&(p1>>i)&1)d++;
42         else if(d&1)return 0;
43     return 1;
44 }
45 void dfs(int p){
46     if(!id[p])id[p]=++ids;
47     for(int i=0;i<256;i++){
48         if(cango(p,i)){
49             if(id[i]==0)dfs(i);
50             dmat[id[p]][id[i]]=1;
51         }
52     }
53 }
54 void init(){
55     ids=0;
56     dfs(255);
57     dmat[id[255]][id[255]]=2;
58     m.init(ids,2);
59 }
60 int main(){
61     init();
62     scanf("%d",&cas);
63     for(int ca=1;ca<=cas;ca++){
64         scanf("%d",&n);
65         printf("Case %d: %d\n",ca,m.binMat(n).mz[1][1]);
66     }
67 }