// #pragma GCC optimize("O3") //调试时关闭

#include<bits/stdc++.h>
using namespace std;

/*----------Consts----------*/
const double eps=1e-6;

const double pi = acos(-1.0);
const long long INF=0x3fffffffffffffff;
/*----------Consts----------*/

typedef long long ll;
typedef long double ld;
typedef pair<ll,ll> pll;

namespace DEFINITION
{
    #define scanfll(a) scanf("%lld",&a)
    #define lowbit(x) ((x)&(-(x)))
    #define ALL(A) (A).begin(),(A).end()
    #define SORT(A) sort(ALL(A))
    #define SORT_REV(A) sort((A).rbegin(),(A).rend())
    //SORT BEFORE UNIQUE!!
    #define UNIQUE(A) A.erase(unique(ALL(A)),(A).end())
    #define Presentation(i,r) " \n"[i==r]
    #define FORLL(i,l,r) for(ll i=(l);i<=(r);i++)
    #define FORLL_rev(i,r,l) for(ll i=(r);i>=(l);i--)
    #define NO cout << "No\n"
    #define YES cout << "Yes\n"
    #define x first
    #define y second
    #define endl '\n' //交互题不启用！
}

namespace CCLIB
{
    #define create_vec(v,n) vector<ll> v(n);for(auto &x:v) cin >> x;
    ostream& operator<<(ostream &out, const pair<ll,ll> &p) {out << '(' << p.first << ',' << p.second << ')';return out;}

    //扩欧返回d=gcd(a,b);x,y对应ax+by=d的解;通解为x=x0+k*b/d,y=y0-k*a/d;
    ll Exgcd(ll a,ll b,ll &x,ll &y) {if(a==0&&b==0) return -1; if(b==0) {x=1;y=0;return a;} ll d=Exgcd(b,a%b,y,x); y-=a/b*x; return d;}

    template<typename T>
    void chmax(T &a,T b){if(a<b) a=b;}
    template<typename T>
    void chmin(T &a,T b){if(a>b) a=b;}
    template<class T> //iterable
    void print_vec(const T &A){for(auto &x:A) cout << x << ' ';cout << endl;}
    void print_float(ld value,int digit=10){cout << fixed << setprecision(digit) << value;}

    vector<ll> snums;
    void Get_Nums(string s) { snums.clear(); ll i=0,cur=0,sgn=1,fl=0; while(i<s.size()) { if(s[i]=='-') sgn=-1; else if(s[i]>='0'&&s[i]<='9') { cur=cur*10+s[i]-'0';fl=1; } else if(fl) { snums.push_back(cur*sgn); cur=0; sgn=1; fl=0; } i++; } if(fl) snums.push_back(cur*sgn); } 

}

namespace MODULE
{
    ll MOD = 1e9+7;
    inline ll Get_Mod(ll x,ll mod=MOD){
        if(x<0) return x-x/mod*mod+mod;
        else return x-x/mod*mod;
    }
    ll qcpow(ll a,ll b,ll p=MOD){ll ret=1;a=Get_Mod(a);for (;b;b>>=1,a=a*a%p) if(b&1) ret=ret*a%p;return ret;}
    ll inv(ll a,ll p=MOD){return qcpow(a,p-2,p);}

    inline ll add(ll a,ll b){return Get_Mod(a+b);}
    inline ll sub(ll a,ll b){return Get_Mod(a-b);}
    inline ll mul(ll a,ll b){return Get_Mod(a*b);}

    void addto(ll &a,ll b){a=Get_Mod(a+b);}
    void subto(ll &a,ll b){a=Get_Mod(a-b);}
    void multo(ll &a,ll b){a=Get_Mod(a*b);}

    vector<ll> Fac;
    void Prepare_Factorial(ll n){Fac.resize(n+1);Fac[0]=1;for(ll i=1;i<=n;i++) Fac[i]=mul(Fac[i-1],i);}
    inline ll C(ll n,ll m){return n<m?0:mul(Fac[n],inv(mul(Fac[m],Fac[n-m])));}
}


using namespace DEFINITION;
using namespace CCLIB;
// using namespace MODULE;

#define ONLINE_JUDGE
#define FAST_IO
#define MUTIPLE_JUDGE

/*----------Code Area----------*/
const ll N = 200005;
void prepare(){
    // Prepare_Factorial(5005);
    // MOD = 1e9+7;
}
vector<vector<ll>> G;
vector<int> vis;
ll n;
int dfs(ll u,ll fa){
    int res=1;
    for(auto v:G[u]){
        if(v==fa) continue;
        if(vis[v]) return 0;
        vis[v]=1;
        res&=dfs(v,u);
    }return res;
}
void solve()
{
    cin >> n;
    G.clear();G.resize(n+1);
    vis.clear();vis.resize(n+1,0);
    ll u,v;
    FORLL(i,1,n){
        cin >> u >> v;
        G[u-1].emplace_back(v);
        G[v].emplace_back(u-1);
    }
    if(dfs(0,-1)) cout << 1 << endl;
    else cout << 0 << endl;
}
/*----------Code Area----------*/

signed main(){

#ifndef ONLINE_JUDGE
    clock_t clk = clock();
    ifstream ifs("1.in");
    ofstream ofs("1.out");
    streambuf *cinbackup;
    cinbackup = cin.rdbuf(ifs.rdbuf());
    streambuf *coutbackup;
    coutbackup = cout.rdbuf(ofs.rdbuf());
#endif

#ifdef FAST_IO
#ifdef ONLINE_JUDGE
    ios::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);
#endif
#endif

long long T = 1;
#ifdef MUTIPLE_JUDGE
    cin >> T;
#endif
prepare();
while(T--) solve();

#ifndef ONLINE_JUDGE
    cin.rdbuf(cinbackup);
    cout.rdbuf(coutbackup);
    cout << clock() - clk << " ms\n";
#endif
    return 0;
}