Fox Ciel is participating in a party in Prime Kingdom. There are n foxes there (include Fox Ciel). The i-th fox is ai years old.
They will have dinner around some round tables. You want to distribute foxes such that:
- Each fox is sitting at some table.
- Each table has at least 3 foxes sitting around it.
- The sum of ages of any two adjacent foxes around each table should be a prime number.
If k foxes f1, f2, ..., fk are sitting around table in clockwise order, then for 1 ≤ i ≤ k - 1: fi and fi + 1 are adjacent, and f1 and fk are also adjacent.
If it is possible to distribute the foxes in the desired manner, find out a way to do that.
Input
The first line contains single integer n (3 ≤ n ≤ 200): the number of foxes in this party.
The second line contains n integers ai (2 ≤ ai ≤ 104).
Output
If it is impossible to do this, output "Impossible".
Otherwise, in the first line output an integer m (): the number of tables.
Then output m lines, each line should start with an integer k -=– the number of foxes around that table, and then k numbers — indices of fox sitting around that table in clockwise order.
If there are several possible arrangements, output any of them.
Sample Input
Input
43 4 8 9
Output
14 1 2 4 3
Input
52 2 2 2 2
Output
Impossible
Input
122 3 4 5 6 7 8 9 10 11 12 13
Output
112 1 2 3 6 5 12 9 8 7 10 11 4
Input
242 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Output
36 1 2 3 6 5 410 7 8 9 12 15 14 13 16 11 108 17 18 23 22 19 20 21 24 建二分图，再用流量为2限制度数，即可用网络流AC。
1 #include <iostream>
2 #include <cstring>
3 #include <cstdio>
4 #include <queue>
5 using namespace std;
6 const int N=210,M=60010,INF=1000000000;
7 int cnt,fir[N],fron[N],nxt[M],to[M],cap[M];
8 int dis[N],gap[N],path[N],vis[N],q[N],h,t;
9 int n,m,a[N],tp[N],tot;vector<int>ans[N];
10 struct Net_Flow{
11 void Init(){
12 memset(fir,0,sizeof(fir));
13 memset(gap,0,sizeof(gap));
14 memset(dis,0,sizeof(dis));
15 h=t=cnt=1;
16 }
17 void addedge(int a,int b,int c){
18 nxt[++cnt]=fir[a];to[fir[a]=cnt]=b;cap[cnt]=c;
19 nxt[++cnt]=fir[b];to[fir[b]=cnt]=a;cap[cnt]=0;
20 }
21 bool BFS(int S,int T){
22 dis[q[h]=T]=1;
23 while(h<=t){
24 int x=q[h++];
25 for(int i=fir[x];i;i=nxt[i])
26 if(!dis[to[i]])dis[q[++t]=to[i]]=dis[x]+1;
27 }
28 return dis[S];
29 }
30 int ISAP(int S,int T){
31 if(!BFS(S,T))return 0;
32 for(int i=S;i<=T;i++)gap[dis[i]]+=1;
33 for(int i=S;i<=T;i++)fron[i]=fir[i];
34 int ret=0,f,p=S,Min;
35 while(dis[S]<=T+1){
36 if(p==T){f=INF;
37 while(p!=S){
38 f=min(f,cap[path[p]]);
39 p=to[path[p]^1];
40 }ret+=f,p=T;
41 while(p!=S){
42 cap[path[p]]-=f;
43 cap[path[p]^1]+=f;
44 p=to[path[p]^1];
45 }
46 }
47 for(int &i=fron[p];i;i=nxt[i])
48 if(cap[i]&&dis[to[i]]==dis[p]-1){
49 path[p=to[i]]=i;break;
50 }
51 if(!fron[p]){
52 if(!--gap[dis[p]])break;Min=T+1;
53 for(int i=fir[p];i;i=nxt[i])
54 if(cap[i])Min=min(Min,dis[to[i]]);
55 gap[dis[p]=Min+1]+=1;fron[p]=fir[p];
56 if(p!=S)p=to[path[p]^1];
57 }
58 }
59 return ret;
60 }
61 void DFS(int x,int id){
62 vis[x]=1;ans[id].push_back(x);
63 for(int i=fir[x];i;i=nxt[i]){
64 if(vis[to[i]]||to[i]<1||to[i]>n)continue;
65 if(a[to[i]]%2==0&&cap[i^1]==0)DFS(to[i],id);
66 if(a[to[i]]%2!=0&&cap[i]==0)DFS(to[i],id);
67 }
68 }
69 void Solve(int S,int T){
70 for(int x=1;x<=n;x++)
71 if(a[x]%2==0&&!vis[x])
72 DFS(x,++tot);
73 printf("%d\n",tot);
74 for(int i=1;i<=tot;i++){
75 printf("%d ",ans[i].size());
76 for(int j=0;j<ans[i].size();j++)
77 printf("%d ",ans[i][j]);
78 puts("");
79 }
80 }
81 }isap;
82 int S,T;
83 bool Check(int x){
84 for(int i=2;i*i<=x;i++)
85 if(x%i==0)return false;
86 return true;
87 }
88 int main(){
89 scanf("%d",&n);isap.Init();T=n+1;
90 if(n%2==1){puts("Impossible");return 0;}
91 for(int i=1;i<=n;i++)scanf("%d",&a[i]);
92 for(int i=1;i<=n;i++){
93 if(a[i]%2==0)isap.addedge(S,i,2);
94 else isap.addedge(i,T,2);
95 }
96 for(int i=1;i<=n;i++)if(a[i]%2==0)
97 for(int j=1;j<=n;j++)if(a[j]%2)
98 if(Check(a[i]+a[j]))
99 isap.addedge(i,j,1);
100 if(isap.ISAP(S,T)!=n){
101 puts("Impossible");
102 return 0;
103 }
104 isap.Solve(S,T);
105 return 0;
106 }