Problem: Note that the only difference between the easy and hard version is the constraint on the number of queries. You can make hacks only if all versions of the problem are solved.

This is an interactive problem.

You are given a tree consisting of n nodes numbered with integers from 1 to n. Ayush and Ashish chose two secret distinct nodes in the tree. You need to find out both the nodes. You can make the following query: 

  * Provide a list of nodes and you will receive a node from that list whose sum of distances to both the hidden nodes is minimal (if there are multiple such nodes in the list, you will receive any one of them). You will also get the sum of distances of that node to the hidden nodes. 



Recall that a tree is a connected graph without cycles. The distance between two nodes is defined as the number of edges in the simple path between them.

More formally, let's define two hidden nodes as s and f. In one query you can provide the set of nodes \\{a_1, a_2, …, a_c\} of the tree. As a result, you will get two numbers a_i and dist(a_i, s) + dist(a_i, f). The node a_i is any node from the provided set, for which the number dist(a_i, s) + dist(a_i, f) is minimal.

You can ask no more than 14 queries.

Input

The first line contains a single integer t (1 ≤ t ≤ 10) — the number of test cases. Please note, how the interaction process is organized.

The first line of each test case consists of a single integer n (2 ≤ n ≤ 1000) — the number of nodes in the tree.

The next n - 1 lines consist of two integers u, v (1 ≤ u, v ≤ n, u ≠ v) — the edges of the tree.

Interaction

To ask a query print a single line: 

  * In the beginning print "? c " (without quotes) where c (1 ≤ c ≤ n) denotes the number of nodes being queried, followed by c distinct integers in the range [1, n] — the indices of nodes from the list. 



For each query, you will receive two integers x, d — the node (among the queried nodes) with the minimum sum of distances to the hidden nodes and the sum of distances from that node to the hidden nodes. If the subset of nodes queried is invalid or you exceeded the number of queries then you will get x = d = -1. In this case, you should terminate the program immediately.

When you have guessed the hidden nodes, print a single line "! " (without quotes), followed by two integers in the range [1, n] — the hidden nodes. You can output the hidden nodes in any order.

After this, you should read a string. If you guess the nodes correctly, you will receive the string "Correct". In this case, you should continue solving the remaining test cases or terminate the program, if all test cases were solved. Otherwise, you will receive the string "Incorrect". In this case, you should terminate the program immediately.

Guessing the hidden nodes does not count towards the number of queries asked.

The interactor is not adaptive. The hidden nodes do not change with queries.

Do not forget to read the string "Correct" / "Incorrect" after guessing the hidden nodes.

You need to solve each test case before receiving the input for the next test case.

The limit of 14 queries applies to each test case and not to the entire input.

After printing a query do not forget to output the end of the line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:

  * fflush(stdout) or cout.flush() in C++;
  * System.out.flush() in Java;
  * flush(output) in Pascal;
  * stdout.flush() in Python;
  * see the documentation for other languages.



Hacks

To hack the solution, use the following test format:

The first line should contain a single integer t (1 ≤ t ≤ 10) — the number of test cases. The description of the test cases follows.

The first line of each test case should contain a single integer n (2 ≤ n ≤ 1000) — the number of nodes in the tree. The second line should contain two distinct integers in the range [1, n] — the hidden nodes. The next n - 1 lines should contain two integers u, v (1 ≤ u, v ≤ n, u ≠ v) — the edges of the tree.

Example

Input


1
3
1 2
1 3

1 1

2 3

3 1

3 1

Correct

Output


? 1 1

? 1 2

? 1 3

? 2 2 3

! 1 3

Note

The tree from the first test is shown below, and the hidden nodes are 1 and 3.

<image>

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
public class EdA {
	static long[] mods = {1000000007, 998244353, 1000000009};
	static long mod = mods[0];
	public static MyScanner sc;
    public static PrintWriter out;
    static int[] depth;
    static ArrayList<ArrayList<Integer>> tree;
	public static void main(String[] omkar) throws Exception{
		// TODO Auto-generated method stub
 		sc = new MyScanner();
 		out = new PrintWriter(System.out);
 		int t = sc.nextInt();
 		for(int i = 0;i<t;i++){
	 		int n = sc.nextInt();
	 		tree = new ArrayList<ArrayList<Integer>>();
	 		for(int j = 0;j<=n;j++)
	 			tree.add(new ArrayList<Integer>());
	 		depth = new int[n+1];
	 		for(int j =0 ;j<n-1;j++){
	 			int v1 = sc.nextInt();
	 			int v2 = sc.nextInt();
	 			tree.get(v1).add(v2);
	 			tree.get(v2).add(v1);
	 		}
	 		out.print("? " + n + " ");
	 		for(int j = 1;j<=n;j++)
	 			out.print(j + " ");
	 		out.println();
	 		out.flush();
	 		int x = sc.nextInt();
	 		int len = sc.nextInt();
	 		dfs(x, -1);
	 		int min = len/2;
	 		int max = len;
	 		while(min < max){
	 			int mid = (min+max+1)/2;
	 			Set<Integer> nodes = new HashSet<Integer>();
	 			for(int j =1;j<=n;j++)
	 				if (depth[j] >= mid && depth[j] <= max)
	 					nodes.add(j);
	 			if (nodes.size() == 0){
	 				max = mid-1;
	 				continue;
	 			}
	 			out.print("? " + nodes.size() + " ");
	 			for(int j : nodes)
	 				out.print(j + " ");
	 			out.println();
	 			out.flush();
	 			int x1 = sc.nextInt();
	 			int len1 = sc.nextInt();
	 			if (len1 == len)
	 				min = mid;
	 			else
	 				max = mid-1;
	 		}
	 		Set<Integer> nodes = new HashSet<Integer>();
			for(int j =1;j<=n;j++)
				if (depth[j] == min)
					nodes.add(j);
			out.print("? " + nodes.size() + " ");
			for(int j : nodes)
				out.print(j + " ");
			out.println();
			out.flush();
			int x1 = sc.nextInt();
			int len1 = sc.nextInt();
			
			Set<Integer> distlen = new HashSet<Integer>();
			depth[x1] = 0;
			dfs(x1, -1);
			for(int j =1;j<=n;j++)
				if (depth[j] == len)
					distlen.add(j);
			out.print("? " + distlen.size() + " ");
			for(int j : distlen)
				out.print(j + " ");
			out.println();
			out.flush();
			int x2 = sc.nextInt();
			int len2 = sc.nextInt();
			out.println("! " + x1 + " " + x2);
			out.flush();
			String verdict = sc.next();
 		}
 		out.close();	
 		return;
 	}
	public static void dfs(int v, int p){
		for(int child : tree.get(v)){
			if (child == p)
				continue;
			depth[child] = depth[v]+1;
			dfs(child, v);
		}
	}
	public static void sort(int[] array){
		ArrayList<Integer> copy = new ArrayList<Integer>();
		for (int i : array)
			copy.add(i);
		Collections.sort(copy);
		for(int i = 0;i<array.length;i++)
			array[i] = copy.get(i);
	}
	static String[] readArrayString(int n){
		String[] array = new String[n];
		for(int j =0 ;j<n;j++)
			array[j] = sc.next();
		return array;
	}
	static int[] readArrayInt(int n){
    	int[] array = new int[n];
    	for(int j = 0;j<n;j++)
    		array[j] = sc.nextInt();
    	return array;
    }
	static int[] readArrayInt1(int n){
		int[] array = new int[n+1];
		for(int j = 1;j<=n;j++){
			array[j] = sc.nextInt();
		}
		return array;
	}
	static long[] readArrayLong(int n){
		long[] array = new long[n];
		for(int j =0 ;j<n;j++)
			array[j] = sc.nextLong();
		return array;
	}
	static double[] readArrayDouble(int n){
		double[] array = new double[n];
		for(int j =0 ;j<n;j++)
			array[j] = sc.nextDouble();
		return array;
	}
	static int minIndex(int[] array){
		int minValue = Integer.MAX_VALUE;
		int minIndex = -1;
		for(int j = 0;j<array.length;j++){
			if (array[j] < minValue){
				minValue = array[j];
				minIndex = j;
			}
		}
		return minIndex;
	}
	static int minIndex(long[] array){
		long minValue = Long.MAX_VALUE;
		int minIndex = -1;
		for(int j = 0;j<array.length;j++){
			if (array[j] < minValue){
				minValue = array[j];
				minIndex = j;
			}
		}
		return minIndex;
	}
	static int minIndex(double[] array){
		double minValue = Double.MAX_VALUE;
		int minIndex = -1;
		for(int j = 0;j<array.length;j++){
			if (array[j] < minValue){
				minValue = array[j];
				minIndex = j;
			}
		}
		return minIndex;
	}
	static long power(long x, long y){
		if (y == 0)
			return 1;
		if (y%2 == 1)
			return (x*power(x, y-1))%mod;
		return power((x*x)%mod, y/2)%mod;
	}
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;
        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } 
                catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt() {
            return Integer.parseInt(next());
        }
        long nextLong() {
            return Long.parseLong(next());
        }
        double nextDouble() {
            return Double.parseDouble(next());
        }
        String nextLine() {
            String str = "";
            try{
                str = br.readLine();
            } 
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
        
    }	
}

//StringJoiner sj = new StringJoiner(" "); 
//sj.add(strings)
//sj.toString() gives string of those stuff w spaces or whatever that sequence is