#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pl = pair<ll,ll>;
using vl = vector<ll>;
#define mp make_pair
#define f first
#define s second
// vectors
#define sz(x) (int)(x).size()
#define all(x) begin(x), end(x)
#define ins insert
#define pf push_front
#define pb push_back
// loops
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define trav(a,x) for (auto& a: x)
const int MOD = 1e9+7; // 998244353;
const ll INF = 1e18; // not too close to LLONG_MAX
#define int ll
const int xd[4] = {1,0,-1,0}, yd[4] = {0,1,0,-1}; // for every grid problem!!
const int N = 5000+5;
vl g[N],tr[N];
int salary[N];
bool vis[N];
void dfs(int u,int p){
salary[u] = 1;
trav(v,tr[u]){
if(v == p) continue;
dfs(v,u);
salary[u] += salary[v];
}
}
void solve(){
int n;
cin >> n;
FOR(i,1,n+1){
int k;
cin >> k;
FOR(j,1,k+1){
int p;cin >> p;
g[p].pb(i);
}
}
int ans= INF;
FOR(i,1,n+1){
// root the tree at this node and run a dfs
// salary[u] = sum salary[v] + 1
FOR(j,1,n+1){
vis[j] = 0;
salary[j] = -1;
}
queue<int> q;
q.push(i);
vis[i] = 1;
salary[i] = 1;
while(!q.empty()){
int u = q.front();
q.pop();
trav(v,g[u]){
if(vis[v] == 0){
vis[v] = 1;
q.push(v);
salary[v] = salary[u] + 1;
}
}
}
int sum = 0;
bool gg = 1;
FOR(j,1,n+1) sum += salary[j],gg &= (salary[j] > 0);
if(gg)
ans = min(ans,sum);
}
cout << ans << "\n";
}
signed main() {
cin.tie(0)->sync_with_stdio(0);
cout.tie(0);
// you should actually read the stuff at the bottom
int t=1;
// cin >> t;
while(t--){
solve();
}
}
/* stuff you should look for
* read the probem 3 more times in case of WA :)
* int overflow, array bounds
* special cases (n=1?)
* do smth instead of nothing and stay organized
* WRITE STUFF DOWN
* DON'T GET STUCK ON ONE APPROACH
*/
