description: A tree is a connected graph without cycles. A rooted tree has a special vertex called the root. The parent of a vertex v (different from root) is the previous to v vertex on the shortest path from the root to the vertex v. Children of the vertex v are all vertices for which v is the parent.
  
  A vertex is a leaf if it has no children. We call a vertex a bud, if the following three conditions are satisfied: 
  
    * it is not a root, 
    * it has at least one child, and 
    * all its children are leaves. 
  
  
  
  You are given a rooted tree with n vertices. The vertex 1 is the root. In one operation you can choose any bud with all its children (they are leaves) and re-hang them to any other vertex of the tree. By doing that you delete the edge connecting the bud and its parent and add an edge between the bud and the chosen vertex of the tree. The chosen vertex cannot be the bud itself or any of its children. All children of the bud stay connected to the bud.
  
  What is the minimum number of leaves it is possible to get if you can make any number of the above-mentioned operations (possibly zero)?
  
  Input
  
  The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.
  
  The first line of each test case contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of the vertices in the given tree.
  
  Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v) meaning that there is an edge between vertices u and v in the tree.
  
  It is guaranteed that the given graph is a tree.
  
  It is guaranteed that the sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
  
  Output
  
  For each test case print a single integer — the minimal number of leaves that is possible to get after some operations.
  
  Example
  
  Input
  
  
  5
  7
  1 2
  1 3
  1 4
  2 5
  2 6
  4 7
  6
  1 2
  1 3
  2 4
  2 5
  3 6
  2
  1 2
  7
  7 3
  1 5
  1 3
  4 6
  4 7
  2 1
  6
  2 1
  2 3
  4 5
  3 4
  3 6
  
  
  Output
  
  
  2
  2
  1
  2
  1
  
  Note
  
  In the first test case the tree looks as follows:
  
  <image>
  
  Firstly you can choose a bud vertex 4 and re-hang it to vertex 3. After that you can choose a bud vertex 2 and re-hang it to vertex 7. As a result, you will have the following tree with 2 leaves:
  
  <image>
  
  It can be proved that it is the minimal number of leaves possible to get.
  
  In the second test case the tree looks as follows:
  
  <image>
  
  You can choose a bud vertex 3 and re-hang it to vertex 5. As a result, you will have the following tree with 2 leaves:
  
  <image>
  
  It can be proved that it is the minimal number of leaves possible to get.
solution: #include <bits/stdc++.h>
  using namespace std;
  using ll = long long;
  using ull = unsigned long long;
  using ld = long double;
  using pii = pair<int, int>;
  using pll = pair<ll, ll>;
  using qi = queue<int>;
  using stki = stack<int>;
  template <typename T>
  using vec = vector<T>;
  template <typename T>
  using us = unordered_set<T>;
  template <typename T>
  using os = set<T>;
  template <typename T>
  using ms = multiset<T>;
  template <typename T1, typename T2>
  using um = unordered_map<T1, T2>;
  template <typename T1, typename T2>
  using om = map<T1, T2>;
  template <typename T>
  using pq = priority_queue<T>;
  template <typename T>
  using pqg = priority_queue<T, vec<T>, greater<T> >;
  using vi = vec<int>;
  using vpii = vec<pii>;
  using vll = vec<ll>;
  using vpll = vec<pll>;
  using vb = vec<bool>;
  const int INF = 0x3f3f3f3f;
  const ll LLINF = 0x3f3f3f3f3f3f3f3f;
  const int MOD = 1e9 + 7;
  const int dir[8][2] = {{0, 1},  {0, -1}, {-1, 0},  {1, 0},
                         {-1, 1}, {1, -1}, {-1, -1}, {1, 1}};
  struct Edge {
    int u, v, w;
  };
  namespace comfun {
  template <typename T>
  inline void cmax(T& u, T v) {
    u = max(u, v);
  }
  template <typename T>
  inline void cmin(T& u, T v) {
    u = min(u, v);
  }
  template <typename T>
  inline T lowbit(T x) {
    return x & (-x);
  }
  template <typename T>
  inline T gcd(T a, T b) {
    if (a == 0) return b;
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  template <typename T>
  inline T lcm(T a, T b) {
    return a / gcd(a, b) * b;
  }
  template <typename T1, typename T2>
  inline T1 fast_pow(T1 a, T2 b) {
    if (b == 1) return a;
    if (b == 0) return 1;
    T1 tmp = fast_pow(a, b >> 1);
    if (b & 1)
      return tmp * tmp * a;
    else
      return tmp * tmp;
  }
  template <typename T1, typename T2, typename T3>
  inline T1 fast_pow(T1 a, T2 b, T3 m) {
    if (b == 1) return a % m;
    if (b == 0) return 1;
    T1 tmp = fast_pow(a, b >> 1, m);
    if (b & 1)
      return tmp * tmp % m * a % m;
    else
      return tmp * tmp % m;
  }
  template <typename T>
  inline bool is_prime(T x) {
    if (x == 1) return false;
    for (T i = 2; i * i <= x; i++)
      if (x % i == 0) return false;
    return true;
  }
  }  // namespace comfun
  namespace debug {
  template <typename T>
  inline void debugArray(T* arr, int sz) {
    cout << "[";
    for (int i = 0; i < sz; i++) {
      cout << arr[i] << ", ";
    }
    cout << "]\n";
  }
  template <typename T>
  inline void printArray(T* arr, int sz) {
    for (int i = 0; i < sz; i++) {
      cout << arr[i] << " ";
    }
    cout << "\n";
  }
  }  // namespace debug
  namespace fast_io {
  void read(int& number) {
    bool negative = false;
    int c;
    number = 0;
    c = getchar();
    if (c == '-') {
      negative = true;
      c = getchar();
    }
    for (; (c > 47 && c < 58); c = getchar()) number = number * 10 + c - 48;
    if (negative) number *= -1;
  }
  }  // namespace fast_io
  using namespace comfun;
  using namespace debug;
  using namespace fast_io;
  inline void init1() {}
  inline void init2() {}
  void dfs(int u, int fa, vi e[], vi& cnt, vi& cut) {
    if (e[u].size() == 1 && u != 1) {
      cnt[u]++;
      return;
    }
    bool flag = 1;
    for (int i = 0; i < e[u].size(); ++i) {
      int v = e[u][i];
      if (v == fa) continue;
      dfs(v, u, e, cnt, cut);
      if (cnt[v] == 0) flag = 0;
      if (cnt[v] == -1) {
        cut[u]++;
      }
    }
    if ((e[u].size() - cut[u] == 1 && u != 1) ||
        (u == 1 && e[u].size() - cut[u] == 0)) {
      if (cnt[u] > 0) return;
      cnt[u]++;
      return;
    }
    if (flag && u != 1) {
      cnt[u]--;
    }
  }
  inline void solve() {
    int n;
    cin >> n;
    vi e[n + 1], cnt(n + 1), cut(n + 1);
    for (int i = 1; i < n; i++) {
      int u, v;
      cin >> u >> v;
      e[u].push_back(v);
      e[v].push_back(u);
    }
    dfs(1, 0, e, cnt, cut);
    int ans = 0;
    for (int i = 1; i <= n; i++) {
      ans += cnt[i];
    }
    cout << ans << '\n';
  }
  inline void setIO() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  }
  int main() {
    setIO();
    init1();
    srand(time(0));
    int t;
    cin >> t;
    for (int i = 1; i <= t; i++) {
      if (0) cout << "Case #" << i << ": ";
      init2();
      solve();
    }
    string jack = "Jack is always within you";
    return 0;
  }
  