(CODEFORCES) Hemose_in_ICPC.CPP

//Problem Link: https://codeforces.com/contest/1592/problem/D

/*The maximum gcd of a path equals the maximum weight of an edge in the tree. Let x be the value of the maximum weight of an edge in the tree, We need to find u, v such that there's an edge between u and v with weight equals x.
Let's find x by putting all the n nodes in the same query, Now we need to find u, v.
If we have an array of edges such that for any consecutive subarray: The component of nodes inside the subarray is connected using the edges inside the subarray. Then we can binary search on this array to find the edge with the maximum weight.
If we put the edges in the array using the order of Euler tour traversal, the array will satisfy the condition above, and we can solve the problem.
Total number of queries is 1+log(2∗(n−1)).
*/

#include <bits/stdc++.h>
using namespace std;
int main(){
  int n;
  cin >> n;
  vector<vector<int>> E(n);
  for (int i = 0; i < n - 1; i++){
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    E[u].push_back(v);
    E[v].push_back(u);
  }
  vector<int> d(n);
  for (int i = 0; i < n; i++){
    d[i] = E[i].size();
  }
  queue<int> Q;
  for (int i = 0; i < n; i++){
    if (d[i] == 1){
      Q.push(i);
    }
  }
  vector<int> t;
  while (!Q.empty()){
    int v = Q.front();
    Q.pop();
    t.push_back(v);
    for (int w : E[v]){
      d[w]--;
      if (d[w] == 1){
        Q.push(w);
      }
    }
  }
  reverse(t.begin(), t.end());
  int r = t[0];
  vector<int> p(n, -1);
  queue<int> Q2;
  Q2.push(r);
  while (!Q2.empty()){
    int v = Q2.front();
    Q2.pop();
    for (int w : E[v]){
      if (w != p[v]){
        p[w] = v;
        Q2.push(w);
      }
    }
  }
  cout << "? " << n;
  for (int i = 0; i < n; i++){
    cout << ' ' << i + 1;
  }
  cout << endl;
  int M;
  cin >> M;
  int tv = n, fv = 1;
  while (tv - fv > 1){
    int mid = (tv + fv) / 2;
    cout << "? " << mid;
    for (int i = 0; i < mid; i++){
      cout << ' ' << t[i] + 1;
    }
    cout << endl;
    int x;
    cin >> x;
    if (x == M){
      tv = mid;
    } else {
      fv = mid;
    }
  }
  int a = t[tv - 1];
  int b = p[a];
  cout << "! " << a + 1 << ' ' << b + 1 << endl;
}