/* 葉までの距離の最大値をもっておいて全方位? */ #include using namespace std; #define ALL(x) (x).begin(),(x).end() #define REP(i, n) for(ll i=0; i<(ll)(n); i++) template int LB(const vector& v, T x) { return lower_bound(ALL(v),x)-v.begin(); } template int UQ(T& v) { sort(ALL(v)); v.erase(unique(ALL(v)),v.end()); return v.size(); } template bool chmax(T &a, T b) { return a bool chmin(T &a, T b) { return a>b ? a=b, true : false; } template using rpriority_queue = priority_queue,greater>; using ll=long long; const int INF=1e9+10; const ll INFL=4e18; using ld=long double; using lll=__int128_t; using ull=unsigned long long; using VI=vector; using VVI=vector; using VL=vector; using VVL=vector; using PL=pair; using VP=vector; using WG=vector>>; #ifdef LOCAL #include "./debug.hpp" #else #define debug(...) #define print_line #endif //---------------------------------------------------------- void solve() { ll N; cin>>N; VVI G(N); REP(i,N-1) { ll u,v; cin>>u>>v; u--; v--; G[u].push_back(v); G[v].push_back(u); } VL dp(N); auto dfs=[&](auto&& dfs, int now, int pre)-> void { for(int nxt: G[now]) if(nxt!=pre) { dfs(dfs,nxt,now); chmax(dp[now],dp[nxt]); } dp[now]++; }; dfs(dfs,0,-1); ll ans=0; auto dfs2=[&](auto&& dfs2, int now, int pre, ll prp)-> void { ll d=G[now].size(); VL dp2(d); REP(i,d) dp2[i]=G[now][i]==pre?prp:dp[G[now][i]]; auto A=dp2; sort(ALL(A)); reverse(ALL(A)); REP(i,d) chmax(ans,(i+1)*A[i]); VL dpl(d+1,-INF), dpr(d+1,-INF); REP(i,d) dpl[i+1]=max(dpl[i],dp2[i]); for(int i=d-1; i>=0; i--) dpr[i]=max(dpr[i+1],dp2[i]); REP(i,d) if(G[now][i]!=pre) { dfs2(dfs2,G[now][i],now,max(dpl[i],dpr[i+1])+1); } }; dfs2(dfs2,0,-1,0); ans++; cout<>T; while(T--) solve(); }