#include<stdio.h>
#include<queue>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int N=1e6+1;
int head[N],cnt,Ev[2*N],Enext[2*N];
int dp1[N]/*把以 i 为根节点的子树作为主枝*/,dp2[N]/*分叉*/; 
int mmin(int x,int y)
{
    return x<y?x:y;
}
void clean(int n)
{
    cnt=0;
    for(int i=1;i<=n;++i)
    head[i]=-1;
}
void addedge(int u,int v)
{
    Ev[cnt]=v;
    Enext[cnt]=head[u];
    head[u]=cnt++;
}
void TreeDp(int id,int fa)
{
    dp1[id]=dp2[id]=0;
    int sum1=0,sum2=0,m1=0,m2=0,son=0;
    for(int i=head[id];i!=-1;i=Enext[i])
    if(Ev[i]!=fa)
    {
        int v=Ev[i];
        TreeDp(v,id);
        sum1+=dp1[v];
        sum2+=dp2[v];
        if(!son)m1=dp1[v]-dp2[v];
        else
        {
            if(dp1[v]-dp2[v]<=m1)
            {
                m2=m1;
                m1=dp1[v]-dp2[v];
            }
            else if(dp1[v]-dp2[v]<m2)m2=dp1[v]-dp2[v];
        }
        son++;
    }
    if(son)
    {
        if(id==1)
        {
            int ans=mmin(sum2+m1+1,sum2+m1+m2+1);
            printf("%d\n",ans);
            return ;
        }
        dp1[id]=sum2+m1;
        dp2[id]=mmin(sum2+2,dp1[id]+2);
        if(son>1)dp2[id]=mmin(dp2[id],sum2+m1+m2+2);
    }
    else
    {
        dp1[id]=0;
        dp2[id]=2;
    }
}
void solve()
{
    int T,id,n,x,y,sum;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        clean(n);
        sum=id=0;
        for(int i=1;i<n;++i)
        {
            scanf("%d%d",&x,&y);
            addedge(x,y);
            addedge(y,x);
        }
        TreeDp(1,1);
    }
}
int main()
{
    solve();
    return 0;
}