#include<iostream>

 #include<algorithm>

 #include<cmath>

 #include<cstring>

 #include<set>

 #include<string>

 #include<vector>

 #include<stack>

 #include<queue>

 using namespace std;

 typedef long long ll;

 const int maxbit = ;

 const int maxn = 1e4+;

 vector<int> G[maxn];

 int depth[maxn];

 int fa[maxn][maxbit];

 int Log[maxn];

 int N;

 void pre(){

     Log[] = -;

     Log[] = ,Log[] = ;

     for(int i = ;i<maxn;i++) Log[i] = Log[i/] + ;

 }

 void dfs(int cur,int father){//dfs预处理

     depth[cur] = depth[father] + ;//当前结点的深度为父亲结点+1

     fa[cur][] = father;//更新当前结点的父亲结点

     for(int j = ;(<<j)<=N;j++){//倍增更新当前结点的祖先

         fa[cur][j] = fa[fa[cur][j-]][j-];

     }

     for(int i = ;i<G[cur].size() ;i++){

         if(G[cur][i] != father) {//dfs遍历

             dfs(G[cur][i],cur);

         }

     }

 }

 int LCA(int u,int v){

     if(depth[u]<depth[v]) swap(u,v);

     int dist = depth[u] - depth[v];//深度差

     while(depth[u]!=depth[v]){//把较深的结点u倍增到与v高度相等

         u = fa[u][Log[depth[u]-depth[v]]];

     }

     if(u == v) return u;//如果u倍增到v，说明v是u的LCA

     for(int i = Log[depth[u]];i>=;i--){//否则两者同时向上倍增

         if(fa[u][i]!=fa[v][i]){//如果向上倍增的祖先不同，说明是可以继续倍增

             u = fa[u][i];//替换两个结点

             v = fa[v][i];

         }

     }

     return fa[u][];//最终结果为u v向上一层就是LCA

 }

 int main()

 {

     int t;

     pre();

     scanf("%d",&t);

     while(t--){

         scanf("%d",&N);

         int root ;

         int in[maxn];

         for(int i = ;i<maxn;i++){

             G[i].clear() ;

         }

         memset(in,,sizeof(in));

         int u,v;

         for(int i = ;i<N-;i++){

             scanf("%d%d",&u,&v);

             in[v] = ;

             G[u].push_back(v);

             G[v].push_back(u);

         }

         for(int i = ;i<=N;i++){//寻树的根结点

             if(in[i] == ) {

                 root = i;

                 break;

             }

         }

         dfs(root,);

         scanf("%d%d",&u,&v);

         int ans = LCA(u,v);

         printf("%d\n",ans);

     }

     return ;

 }