#include <map>

#include <set>

#include <list>

#include <cmath>

#include <queue>

#include <stack>

#include <cstdio>

#include <vector>

#include <string>

#include <cctype>

#include <complex>

#include <cassert>

#include <utility>

#include <cstring>

#include <cstdlib>

#include <iostream>

#include <algorithm>

using namespace std;

typedef pair<int,int> PII;

typedef long long ll;

#define lson l,m,rt<<1

#define pi acos(-1.0)

#define rson m+1,r,rt<<11

#define All 1,N,1

#define N 50

#define read freopen("in.txt", "r", stdin)

const ll  INFll = 0x3f3f3f3f3f3f3f3fLL;

const int INF= 0x7ffffff;

const int mod =  ;

vector<int>e[N];

int used[N],num[N],n,m,len;

map<int,double>dp[N];

//统计各连通分量的节点数

int dfs(int u){

    used[u]=;

    int total=;

    for(int i=;i<e[u].size();++i){

        if(!used[e[u][i]])

            total+=dfs(e[u][i]);

    }

    return total;

}

//记忆化搜索

double solve(int i,int s){

    if(dp[i].count(s))return dp[i][s];

    int liantong=;

    //当前联通的节点数

    for(int j=;j<len;++j){

        if(s&(<<j))

        liantong+=num[j];

    }

    if(liantong==n)return dp[i][s]=;

    //要选择未联通的点需要的平均天数

    dp[i][s]=1.0*(n-)/(n-liantong);

    //选择一个未连接的联通分量

    for(int j=;j<len;++j){

        if(!(s&(<<j)))

        dp[i][s]+=solve(j,s|(<<j))*num[j]/(n-liantong);

    }

    return dp[i][s];

}

int main()

{

    int t,cas=;

    scanf("%d",&t);

    while(t--){

        scanf("%d%d",&n,&m);

        int u,v;

        memset(used,,sizeof(used));

        memset(num,,sizeof(num));

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

            e[i].clear();

        while(m--){

            scanf("%d%d",&u,&v);

            e[u].push_back(v);

            e[v].push_back(u);

        }

        len=;

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

            if(used[i])continue;

            dp[len].clear();

            num[len++]=dfs(i);

        }

        printf("Case %d: %.6lf\n", ++cas, solve(, ));

    }

return ;

}