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.
#include <bits/stdc++.h>
#define elif else if
#define SZ(x) (int)(x).size()
#define ALL(a) a.begin(), a.end()
using namespace std;
using LL = long long;
using LD = long double;
using PII = pair<int, int>;
using PLI = pair<long long, int>;
using PLL = pair<long long, long long>;
using TI = tuple<int, int, int>;
mt19937 mrand(random_device{}());
int rnd(int x) { return mrand() % x; }
template <class T> int bcnt(const T &a) { return __builtin_popcount(a); }
template <> int bcnt(const int64_t &a) { return __builtin_popcount(a); }
template <class T> int blen(const T &a) { return a ? 1 + blen(a / 2) : 0; }
template <class A, class B> B conv(const A &a, B b = B()) {
    stringstream s;
    s << a, s >> b;
    return b;
}
template <class T> T inf() { return numeric_limits<T>::max() / 2; }
template <class T> T sign(const T &a) { return a == 0 ? 0 : (a < 0 ? -1 : 1); }
template <class T> T floor(const T &a, const T &b) {
    assert(b != 0);
    return sign(a) * sign(b) > 0 ? abs(a) / abs(b) : -(abs(a) + abs(b) - 1) / abs(b);
}
template <class T> T ceil(const T &a, const T &b) {
    assert(b != 0);
    return sign(a) * sign(b) > 0 ? (abs(a) + abs(b) - 1) / abs(b) : -abs(a) / abs(b);
}
template <class T> bool chmin(T &a, T b) { return b < a ? a = b, true : false; }
template <class T> bool chmax(T &a, T b) { return b > a ? a = b, true : false; }
template <class T> auto min(const T &a) { return *min_element(ALL(a)); } 
template <class T> auto max(const T &a) { return *max_element(ALL(a)); }
template <class T> auto sum(const T &a) { return accumulate(ALL(a), (typename T::value_type)0); }
template <> auto sum(const vector<int> &a) { return accumulate(ALL(a), 0LL); }
template <> auto sum(const vector<string> &a) { return accumulate(ALL(a), string()); }
template <class T> void uniq(T &a) { sort(ALL(a)); a.erase(unique(ALL(a)), a.end()); }
template <class T> auto vect(const T &v, int n) { return vector<T>(n, v); }
template <class T, class... D> auto vect(const T& v, int n, D... m) {
    return vector<decltype(vect(v, m...))>(n, vect(v, m...));
}
template <class T> void in(T &a) { cin >> a; }
template <class A, class... B> void in(A &a, B &... b) { cin >> a, in(b...); }
template <class T> void in(vector<T> &vec) { for (auto& v: vec) cin >> v; }
void ou() { cout << endl; }
template<class T> void ou(const T &a) { cout << a << endl; }
template <class A, class... B> void ou(const A &a, const B &... b) {
    cout << a << ' ', ou(b...);
}
template <class T> void ou(const vector<T> &vec) {
    cout << (int)vec.size() << ' ';
    bool first = true;
    for (auto& v: vec) { cout << (first ? "" : " "); cout << v; first = 0; }
    cout << '\n';
}
 
template<class A, class B> ostream& operator <<(ostream& out, const pair<A, B> &p) {
    return out << "(" << p.first << ", " << p.second << ")";
}
template<class A, class B, class C> ostream& operator <<(ostream& out, const tuple<A, B, C> &p) {
    return out << "(" << get<0>(p) << ", " << get<1>(p) << ", " << get<2>(p) << ")";
}
template <typename T, size_t N>
ostream& operator <<(ostream& out, const array<T, N>& a) {
    out << "["; bool first = true;
    for (auto& v : a) { out << (first ? "" : ", "); out << v; first = 0;} out << "]";
    return out;
}
template <typename T>
ostream& operator <<(ostream& out, const vector<T>& a) {
    out << "["; bool first = true;
    for (auto& v : a) { out << (first ? "" : ", "); out << v; first = 0;} out << "]";
    return out;
}
template <typename T>
ostream& operator <<(ostream& out, const deque<T>& a) {
    out << "["; bool first = true;
    for (auto& v : a) { out << (first ? "" : ", "); out << v; first = 0;} out << "]";
    return out;
}
template <typename T, class Cmp>
ostream& operator <<(ostream& out, const set<T, Cmp>& a) {
    out << "{"; bool first = true;
    for (auto& v : a) { out << (first ? "" : ", "); out << v; first = 0;} out << "}";
    return out;
}
template <typename T, class Cmp>
ostream& operator <<(ostream &out, const multiset<T, Cmp>& a) {
    out << "{"; bool first = true;
    for (auto& v : a) { out << (first ? "" : ", "); out << v; first = 0;} out << "}";
    return out;
}
template <typename T>
ostream& operator <<(ostream& out, const unordered_set<T>& a) {
    out << "{"; bool first = true;
    for (auto& v : a) { out << (first ? "" : ", "); out << v; first = 0;} out << "}";
    return out;
}
template <typename U, typename T, class Cmp>
ostream& operator <<(ostream& out, const map<U, T, Cmp>& a) {
    out << "{"; bool first = true;
    for (auto& p : a) { out << (first ? "" : ", "); out << p.first << ":" << p.second; first = 0;} out << "}";
    return out;
}
template<class T> inline void YES(T condition) { 
    if(condition) cout << "YES" << '\n';
    else cout << "NO" << '\n';
}
template<class T> inline void Yes(T condition) {
    if(condition) cout << "Yes" << '\n';
    else cout << "No" << '\n';
}
template<class T> T gcd(T a, T b) {
    while (b) {
        T tmp = b;
        b = a % b;
        a = tmp;
    }
    return a;
}
template<class T> T lcm(T a, T b){
    return a / gcd(a, b) * b;
}
long long qpow(long long a, long long b, int MOD) {
    if (b == 0) return 1;
    long long res = 1;
    while (b) {
        if (b & 1) res = (res * a) % MOD;
        a = (a * a) % MOD;
        b >>= 1;
    }
    return res;
}
 
const int MOD = 1e9 + 7;
const int maxn = 2e6;
const LD eps = 1e-8;
const LD pi = acos(-1);
// const int INF = 2e9;
int fac[maxn + 5];
int invfac[maxn + 5];
void init() {
    fac[0] = 1;
    for (int i = 1; i <= maxn; i++) fac[i] = 1LL * fac[i - 1] * i % MOD;
    invfac[maxn] = (int)(qpow(fac[maxn], MOD - 2, MOD));
    for (int i = maxn - 1; i >= 0; i--) invfac[i] = 1LL * invfac[i + 1] * (i + 1) % MOD;
}
int C(int n, int k) {
    assert(0 <= k && n >= k);
    int tmp = (int)(1LL * invfac[k] * invfac[n - k] % MOD);
    return (int)(1LL * fac[n] * tmp % MOD);
}
void Add(int &a, int b) {
    a += b;
    if (a >= MOD) a -= MOD;
    if (a < 0) a += MOD;
}

// #define FANZHH

#ifdef FANZHH
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
    cerr << name << ": " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
    const char* comma = strchr(names + 1, ',');
    cerr.write(names, comma - names) << ": " << arg1 << " |";
    __f(comma + 1, args...);
}
#else
#define trace(...) (void(0))
#endif

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(20);

    int T; cin >> T;
    while (T--) {
        int n; cin >> n;
        vector<vector<int>> G(n);
        for (int i = 0; i < n - 1; i++) {
            int u, v; cin >> u >> v;
            u--; v--;
            G[u].push_back(v);
            G[v].push_back(u);
        }
        vector<int> tp(n);
        vector<int> isleaf(n, 1);
        vector<int> goodbud(n);
        // find leaves and buds
        function<int(int, int)> dfs = [&](int u, int p) -> int {
            int soncnt = 0;
            for (int& v: G[u]) if (v != p) {
                if (dfs(v, u) == 0) tp[u] = 1;
                isleaf[u] = 0;
                soncnt++;
            }
            if (soncnt > 1) {
                for (int &v: G[u]) if (v != p && tp[v] == 1) goodbud[v] = 1;
            }
            return tp[u];
        }; dfs(0, -1);
        int buds = sum(goodbud);
        int leaves = sum(isleaf);
        trace(tp, isleaf, goodbud, buds, leaves);
        int mx = 0;
        // fix a leaf and maximize buds
        int fixbud = 0;
        function<void(int, int)> dfs2 = [&](int u, int p) {
            fixbud += goodbud[u];
            if (isleaf[u]) {
                chmax(mx, buds - fixbud);
            }
            for (int &v: G[u]) if (v != p) {
                dfs2(v, u);
            }
            fixbud -= goodbud[u];
        }; dfs2(0, -1);
        cout << leaves - mx << '\n';
    }

    return 0;
}
