/*
hdu 5521 最短路

problem:
给你n个点的图,甲在1,乙在n. 它们同时移动,问相聚时的最小花费. 然后是m个点集,点集内的任意两点之间的移动花费为ti

solve:
因为是同时移动. 所以分别对1和n求一个最短路. 然后每个节点取两个最短路中的最大值就能得到花费.
最开始想的是建立所有边,但是边的数量会太多.
所有走到一个节点时,将其所在的所有点集处理一遍. 而且只需要处理一次即可,已经维护了一个最短状态.

hhh-2016-08-30 19:56:48
*/
#pragma comment(linker,"/STACK:124000000,124000000")
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <vector>
#include <math.h>
#include <queue>
#include <set>
#include <map>
#define lson  i<<1
#define rson  i<<1|1
#define ll long long
#define clr(a,b) memset(a,b,sizeof(a))
#define scanfi(a) scanf("%d",&a)
#define scanfs(a) scanf("%s",a)
#define scanfl(a) scanf("%I64d",&a)
#define key_val ch[ch[root][1]][0]
#define inf 0x3f3f3f3f3f3f3f3f
using namespace std;
const ll mod = 1e9+7;
const int maxn = 100010;

struct qnode
{
    int v,c;
    qnode(int _v = 0 ,int _c =0) : v(_v),c(_c){}
    bool operator <(const qnode &a)const
    {
        return c > a.c;
    }
};

bool vis[maxn];
bool tvis[maxn];
ll dis1[maxn], dis2[maxn],cost[maxn];
int anspos[maxn];
int num[maxn];
vector<int> pos[maxn];
vector<int> have[maxn];
int T,n,m;
void dijkstra(int start,ll dis[])
{
    for(int i= 1;i <= n;i++)
    {
        vis[i] = tvis[i] = 0;
        dis[i] = inf;
    }
    priority_queue<qnode> q;
    q.push(qnode(start,0));
    dis[start] = 0;
    qnode t;
    while(!q.empty())
    {
        t = q.top();
        q.pop();
        int u = t.v;
        if(vis[u]) continue;
        vis[u] = 1;
        for(int i = 0 ;i < pos[u].size();i++)
        {
            int t = pos[u][i];
//            if(tvis[t]) continue;
//            tvis[t] = 1;
            for(int j = 0;j<have[t].size();j++)
            {
                int v = have[t][j];
                if(v == u)
                    continue;
                if(dis[v] > dis[u] + cost[t])
                {
                    dis[v] = dis[u] + cost[t];
                    q.push(qnode(v,dis[v]));
                }
            }
        }
    }
}

int main()
{
    int x;
//    freopen("in.txt","r",stdin);
    scanfi(T);
    int cas = 1;
    while(T--)
    {
        scanfi(n),scanfi(m);
        for(int i = 0 ;i <= n;i++)
           pos[i].clear();
        for(int i = 0;i <= m;i++)
            have[i].clear();
        for(int i = 1;i <= m;i++)
        {
            scanf("%I64d%d",&cost[i],&num[i]);
            for(int j = 0;j < num[i];j++)
            {
                scanfi(x);
                pos[x].push_back(i);
                have[i].push_back(x);
            }
        }
        dijkstra(1,dis1);
        dijkstra(n,dis2);
        int cnt = 0;
        ll ans = inf;
        for(int i = 1;i <= n;i++)
        {
            ll t = max(dis1[i],dis2[i]);
            if(t != inf){
                ans = min(ans,t);
            }
        }
        for(int i =1; i <= n;i++)
        {
            if(max(dis1[i],dis2[i]) == ans)
            anspos[cnt++] = i;
        }
        printf("Case #%d: ",cas++);
       if(ans != inf)
       {
           printf("%I64d\n",ans);
           for(int i = 0;i < cnt;i++)
           {
               printf("%d%c",anspos[i],i == cnt-1 ? ‘\n‘:‘ ‘);
           }
       }
       else
        printf("Evil John\n");
    }
    return 0;
}