#include using namespace std; using ll=long long; using ld=long double; const ll ILL=2167167167167167167; const int INF=2100000000; #define rep(i,a,b) for (int i=(int)(a);i<(int)(b);i++) #define all(p) p.begin(),p.end() template using _pq = priority_queue, greater>; template int LB(vector &v,T a){return lower_bound(v.begin(),v.end(),a)-v.begin();} template int UB(vector &v,T a){return upper_bound(v.begin(),v.end(),a)-v.begin();} template bool chmin(T &a,T b){if(b bool chmax(T &a,T b){if(a void So(vector &v) {sort(v.begin(),v.end());} template void Sore(vector &v) {sort(v.begin(),v.end(),[](T x,T y){return x>y;});} bool yneos(bool a,bool upp=false){if(a){cout<<(upp?"YES\n":"Yes\n");}else{cout<<(upp?"NO\n":"No\n");}return a;} template void vec_out(vector &p,int ty=0){ if(ty==2){cout<<'{';for(int i=0;i<(int)p.size();i++){if(i){cout<<",";}cout<<'"'< T vec_min(vector &a){assert(!a.empty());T ans=a[0];for(auto &x:a) chmin(ans,x);return ans;} template T vec_max(vector &a){assert(!a.empty());T ans=a[0];for(auto &x:a) chmax(ans,x);return ans;} template T vec_sum(vector &a){T ans=T(0);for(auto &x:a) ans+=x;return ans;} int pop_count(long long a){int res=0;while(a){res+=(a&1),a>>=1;}return res;} template T square(T a){return a * a;} using len_type=ll; struct edge{ int to; len_type length; }; len_type dijkstra_movement(edge e,int from,len_type t){ return t+e.length; } vector dijkstra(vector> &G,int start,len_type e, len_type inf){ int M=G.size(); assert(M>start&&start>=0); vector seen(M,inf); seen[start]=e; priority_queue,vector>,greater>> pq; pq.push({e,start}); while(!pq.empty()){ len_type time; int ver; tie(time,ver)=pq.top(); pq.pop(); if(seen[ver]> t; rep(i, 0, t) solve(); } void solve(){ int N, M; cin >> N >> M; vector> G(N + M * 2); rep(i, 0, M){ int K; ll C; cin >> K >> C; G[N + i * 2].push_back({N + i * 2 + 1, 1}); G[N + i * 2 + 1].push_back({N + i * 2, 0}); C++; rep(j, 0, K){ int s; cin >> s; s--; if (s & 1){ G[s].push_back({N + i * 2 + 1, s / 2 + 1 + C}); G[N + i * 2 + 1].push_back({s, s / 2}); } else{ G[s].push_back({N + i * 2, s / 2 + C}); G[N + i * 2].push_back({s, s / 2 }); } } } auto ans = dijkstra(G, 0, 0, ILL); if (ans[N - 1] == ILL) cout << "-1\n"; else cout << ans[N - 1] << "\n"; }