#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define f first
#define s second
#define all(x) begin(x), end(x)
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) (int)(x).size()
#define FOR(i, a, b) for(int i = (a); i < (b); i++)
#define F0R(i, x) FOR(i, 0, x)
#define FORd(i, a, b) for(int i = (b)-1; i >= (a); i--)
#define F0Rd(i, x) FORd(i, 0, x)
#define ckif(a, b, c) ((c) ? (a) : (b))
const int MAX_N = 5001;
const ll MOD = 1000000007;
const ll INF = 1e18;
int n;
vector<int> adj[MAX_N];
int main(int argc, const char * argv[]){
ios_base::sync_with_stdio(0), cin.tie(0);
cin >> n;
F0R(i, n){
int a; cin >> a;
F0R(j, a){
int x; cin >> x; x--;
adj[x].push_back(i);
}
}
ll ans = INF;
F0R(i, n){
// i is the root of the tree
// ans for this root is just the sum of the depths of all nodes
// it is optimal to use the shortest path tree starting from i
vector<ll> dist(n, -1);
queue<pll> q; q.push({i, 1});
while(!q.empty()){
ll curr = q.front().f, cost = q.front().s; q.pop();
if(~dist[curr]) continue;
dist[curr] = cost;
for(int child : adj[curr])
q.push({child, cost+1});
}
ll tot = 0;
bool ok = true;
F0R(j, n) tot += dist[j], ok = ok && ~dist[j];
if(ok) ans = min(ans, tot);
}
cout << ans << endl;
return 0;
}
