网络流（最大流）CodeForces 512C：Fox And Dinner

Fox Ciel is participating in a party in Prime Kingdom. There are n foxes there (include Fox Ciel). The i-th fox is a_i 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, ..., f_k are sitting around table in clockwise order, then for 1 ≤ i ≤ k - 1: f_i and f_i + 1 are adjacent, and f₁ and f_k 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 a_i (2 ≤ a_i ≤ 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

4
3 4 8 9

Output

1
4 1 2 4 3

Input

5
2 2 2 2 2

Output

Impossible

Input

12
2 3 4 5 6 7 8 9 10 11 12 13

Output

1
12 1 2 3 6 5 12 9 8 7 10 11 4

Input

24
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

Output

3
6 1 2 3 6 5 4
10 7 8 9 12 15 14 13 16 11 10
8 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。

 #include <iostream>

 #include <cstring>

 #include <cstdio>

 #include <queue>

 using namespace std;

 const int N=,M=,INF=;

 int cnt,fir[N],fron[N],nxt[M],to[M],cap[M];

 int dis[N],gap[N],path[N],vis[N],q[N],h,t;

 int n,m,a[N],tp[N],tot;vector<int>ans[N];

 struct Net_Flow{

     void Init(){

         memset(fir,,sizeof(fir));

         memset(gap,,sizeof(gap));

         memset(dis,,sizeof(dis));

         h=t=cnt=;

     }

     void addedge(int a,int b,int c){

         nxt[++cnt]=fir[a];to[fir[a]=cnt]=b;cap[cnt]=c;

         nxt[++cnt]=fir[b];to[fir[b]=cnt]=a;cap[cnt]=;

     }

     bool BFS(int S,int T){

         dis[q[h]=T]=;

         while(h<=t){

             int x=q[h++];

             for(int i=fir[x];i;i=nxt[i])

                 if(!dis[to[i]])dis[q[++t]=to[i]]=dis[x]+;

         }

         return dis[S];

     }

     int ISAP(int S,int T){

         if(!BFS(S,T))return ;

         for(int i=S;i<=T;i++)gap[dis[i]]+=;

         for(int i=S;i<=T;i++)fron[i]=fir[i];

         int ret=,f,p=S,Min;

         while(dis[S]<=T+){

             if(p==T){f=INF;

                 while(p!=S){

                     f=min(f,cap[path[p]]);

                     p=to[path[p]^];

                 }ret+=f,p=T;

                 while(p!=S){

                     cap[path[p]]-=f;

                     cap[path[p]^]+=f;

                     p=to[path[p]^];

                 }

             }

             for(int &i=fron[p];i;i=nxt[i])

                 if(cap[i]&&dis[to[i]]==dis[p]-){

                     path[p=to[i]]=i;break;

                 }

             if(!fron[p]){

                 if(!--gap[dis[p]])break;Min=T+;

                 for(int i=fir[p];i;i=nxt[i])

                     if(cap[i])Min=min(Min,dis[to[i]]);

                 gap[dis[p]=Min+]+=;fron[p]=fir[p];

                 if(p!=S)p=to[path[p]^];

             }

         }

         return ret;

     }

     void DFS(int x,int id){

         vis[x]=;ans[id].push_back(x);

         for(int i=fir[x];i;i=nxt[i]){

             if(vis[to[i]]||to[i]<||to[i]>n)continue;

             if(a[to[i]]%==&&cap[i^]==)DFS(to[i],id);

             if(a[to[i]]%!=&&cap[i]==)DFS(to[i],id);

         }

     }

     void Solve(int S,int T){

         for(int x=;x<=n;x++)

             if(a[x]%==&&!vis[x])

                 DFS(x,++tot);

         printf("%d\n",tot);

         for(int i=;i<=tot;i++){

             printf("%d ",ans[i].size());

             for(int j=;j<ans[i].size();j++)

                 printf("%d ",ans[i][j]);

             puts("");

         }

     }

 }isap;

 int S,T;

 bool Check(int x){

     for(int i=;i*i<=x;i++)

         if(x%i==)return false;

     return true;

 }

 int main(){

     scanf("%d",&n);isap.Init();T=n+;

     if(n%==){puts("Impossible");return ;}

     for(int i=;i<=n;i++)scanf("%d",&a[i]);

     for(int i=;i<=n;i++){

         if(a[i]%==)isap.addedge(S,i,);

         else isap.addedge(i,T,);

     }

     for(int i=;i<=n;i++)if(a[i]%==)

         for(int j=;j<=n;j++)if(a[j]%)

             if(Check(a[i]+a[j]))

                 isap.addedge(i,j,);

     if(isap.ISAP(S,T)!=n){

         puts("Impossible");

         return ;

     }

     isap.Solve(S,T);

     return ;

 }