#include <bits/stdc++.h>
using namespace std;
template <class T>
void chmin(T &a, const T &b) {
  if (a > b) a = b;
}
template <class T>
void chmax(T &a, const T &b) {
  if (a < b) a = b;
}
template <class T>
using MaxHeap = priority_queue<T>;
template <class T>
using MinHeap = priority_queue<T, vector<T>, greater<T>>;
template <class T>
vector<T> vect(int len, T elem) {
  return vector<T>(len, elem);
}
template <class T, class U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
  os << p.first << ',' << p.second;
  return os;
}
template <class T, class U>
istream &operator>>(istream &is, pair<T, U> &p) {
  is >> p.first >> p.second;
  return is;
}
template <class T>
ostream &operator<<(ostream &os, const vector<T> &vec) {
  for (int i = 0; i < vec.size(); i++) {
    os << vec[i];
    if (i + 1 < vec.size()) os << ' ';
  }
  return os;
}
template <class T>
istream &operator>>(istream &is, vector<T> &vec) {
  for (int i = 0; i < vec.size(); i++) is >> vec[i];
  return is;
}
void fastio() {
  cin.tie(0);
  ios::sync_with_stdio(0);
  cout << fixed << setprecision(20);
}
template <class Cost = int>
struct edge {
  int from, to;
  Cost cost;
  edge() {}
  edge(int f, int t, Cost cost) : from(f), to(t), cost(cost) {}
};
template <class Cost = int>
struct Tree {
  int V;
  int LOG_V;
  int root;
  vector<int> depth;
  vector<vector<int>> dpar;
  vector<int> par;
  vector<int> subtree_size;
  vector<Cost> dist;
  vector<vector<edge<Cost>>> G;
  Tree() {}
  Tree(int V, int r = 0) : V(V), root(r) {
    G.resize(V);
    depth.resize(V);
    par.resize(V);
    dist.resize(V);
    subtree_size.resize(V);
  }
  void add_edge(int u, int v, Cost c = Cost(0)) {
    G[u].push_back(edge<Cost>(u, v, c));
    G[v].push_back(edge<Cost>(v, u, c));
  }
  int dfs(int v, int p, int d, Cost c) {
    depth[v] = d;
    par[v] = p;
    subtree_size[v] = 1;
    dist[v] = c;
    bool is_centroid = true;
    for (auto e : G[v]) {
      int u = e.to;
      if (u == p) continue;
      subtree_size[v] += dfs(u, v, d + 1, c + e.cost);
    }
    return subtree_size[v];
  }
  void dfs() { dfs(root, -1, 0, Cost(0)); }
  void construct_dpar() {
    LOG_V = 1;
    while ((1 << LOG_V) <= V) LOG_V++;
    dpar.resize(LOG_V);
    for (int i = 0; i < LOG_V; i++) dpar[i].resize(V, -1);
    for (int i = 0; i < V; i++) dpar[0][i] = par[i];
    for (int i = 1; i < LOG_V; i++) {
      for (int j = 0; j < V; j++) {
        if (dpar[i - 1][j] == -1)
          dpar[i][j] = -1;
        else
          dpar[i][j] = dpar[i - 1][dpar[i - 1][j]];
      }
    }
    return;
  }
  int kth_parent(int v, int k) {
    int res = v;
    for (int i = 0; i < LOG_V; i++) {
      if (k & (1 << i)) res = dpar[i][res];
      if (res == -1) return -1;
    }
    return res;
  }
  int lca(int u, int v) {
    if (depth[u] > depth[v]) swap(u, v);
    v = kth_parent(v, depth[v] - depth[u]);
    if (u == v) return v;
    for (int i = LOG_V - 1; i >= 0; i--) {
      if (dpar[i][u] != dpar[i][v]) {
        u = dpar[i][u];
        v = dpar[i][v];
      }
    }
    return par[v];
  }
  Cost distance(int u, int v) {
    int l = lca(u, v);
    return dist[u] + dist[v] - dist[l] - dist[l];
  }
  int size() { return V; }
  void dump_tree() {
    for (int i = 0; i < V; i++) {
      printf("id: %d, par %d, depth %d\n", i, par[i], depth[i]);
    }
    return;
  }
};
void solve() {
  int n;
  cin >> n;
  n *= 2;
  Tree<long long> T(n);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    long long c;
    cin >> a >> b >> c;
    a--;
    b--;
    T.add_edge(a, b, c);
  }
  T.dfs();
  function<pair<long long, long long>(int, int)> dfs = [&](int v, int p) {
    pair<long long, long long> res({0ll, 0ll});
    for (const auto &e : T.G[v]) {
      if (e.to == p) continue;
      pair<long long, long long> ch = dfs(e.to, v);
      res.first += ch.first;
      res.second += ch.second;
      if (T.subtree_size[e.to] % 2 == 1) res.first += e.cost;
      res.second +=
          e.cost * min(T.subtree_size[e.to], (n - T.subtree_size[e.to]));
    }
    return res;
  };
  auto ans = dfs(0, -1);
  cout << ans.first << ' ' << ans.second << "\n";
  return;
}
signed main() {
  fastio();
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}
