#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
#define int long long
#define endl "\n"

vector<int> g[N]; // 邻接表，存图
int sz[N];        // sz[i]:以i为根的子树中节点个数
int son[N];       // son[i]:去掉节点i后，剩下的连通分量中最大子树节点个数
int r1, r2, n;

void dfs(int u, int fa) {
    sz[u] = 1;  // u为根的子树中，最起码有一个节点u
    son[u] = 0; // 把节点u去掉后，剩下的连通分量中最大子树节点个数现在还不知道，预求最大，先设最小

    for (int i = 0; i < g[u].size(); i++) { // 枚举u的每一条出边
        int v = g[u][i];
        if (v == fa) continue;
        dfs(v, u);
        sz[u] += sz[v];
        son[u] = max(son[u], sz[v]);
    }
    son[u] = max(son[u], n - sz[u]);
    if ((son[u] << 1) <= n) r2 = r1, r1 = u;
}

signed main() {
    int T;
    cin >> T;
    while (T--) {
        cin >> n;
        // 多组测试数据，清空
        memset(sz, 0, sizeof sz);
        memset(son, 0, sizeof son);
        r1 = r2 = 0;

        for (int i = 0; i <= n + 10; i++) g[i].clear();
        for (int i = 1; i < n; i++) { // n-1条边
            int x, y;
            cin >> x >> y;
            g[x].push_back(y);
            g[y].push_back(x);
        }

        dfs(1, 0); // 以1号点为入口，它的父节点是0

        if (!r2) {
            int r3 = g[r1][0];
            cout << r1 << " " << r3 << endl;
            cout << r1 << " " << r3 << endl;
        } else {
            int r3 = r1;
            for (int i = 0; i < g[r2].size(); i++) {
                r3 = g[r2][i];
                if (r3 != r1) break;
            }
            cout << r3 << " " << r2 << endl;
            cout << r3 << " " << r1 << endl;
        }
    }
    return 0;
}