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.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Scanner;
import java.util.Set;


public class Round646 {

	static class Reader {
		final private int BUFFER_SIZE = 1 << 16;
		private DataInputStream din;
		private byte[] buffer;
		private int bufferPointer, bytesRead;

		public Reader() {
			din = new DataInputStream(System.in);
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}

		public Reader(String file_name) throws IOException {
			din = new DataInputStream(new FileInputStream(file_name));
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}

		public String readLine() throws IOException {
			byte[] buf = new byte[64]; // line length
			int cnt = 0, c;
			while ((c = read()) != -1) {
				if (c == '\n')
					break;
				buf[cnt++] = (byte) c;
			}
			return new String(buf, 0, cnt);
		}

		public int nextInt() throws IOException {
			int ret = 0;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();
			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');

			if (neg)
				return -ret;
			return ret;
		}

		public long nextLong() throws IOException {
			long ret = 0;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();
			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');
			if (neg)
				return -ret;
			return ret;
		}

		public double nextDouble() throws IOException {
			double ret = 0, div = 1;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();

			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');

			if (c == '.') {
				while ((c = read()) >= '0' && c <= '9') {
					ret += (c - '0') / (div *= 10);
				}
			}

			if (neg)
				return -ret;
			return ret;
		}

		private void fillBuffer() throws IOException {
			bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
			if (bytesRead == -1)
				buffer[0] = -1;
		}

		private byte read() throws IOException {
			if (bufferPointer == bytesRead)
				fillBuffer();
			return buffer[bufferPointer++];
		}

		public void close() throws IOException {
			if (din == null)
				return;
			din.close();
		}
	}

	public static long[] initArray(int n, Reader scan) throws IOException {

		long arr[] = new long[n];
		for (int i = 0; i < n; i++) {
			arr[i] = scan.nextLong();
		}

		return arr;
	}

	public static long sum(long arr[]) {

		long sum = 0;
		for (long i : arr) {
			sum += (long) i;
		}

		return sum;
	}

	public static long max(long arr[]) {

		long max = Long.MIN_VALUE;

		for (long i : arr) {
			max = Math.max(i, max);
		}

		return max;
	}

	public static long min(long arr[]) {
		long min = Long.MAX_VALUE;

		for (long i : arr) {
			min = Math.min(i, min);
		}

		return min;
	}
	
	public static List<Integer>[] initAdjacency(int n, int e, Reader scan, boolean type) throws IOException{
		
		List<Integer> adj[]=new ArrayList[n+1];
		
		
		for(int i=0;i<e;i++) {
			int u=scan.nextInt();
			int v=scan.nextInt();
			
			if(adj[u]==null) adj[u]=new ArrayList<>();
			if(type&&adj[v]==null) adj[v]=new ArrayList<>();
			
			adj[u].add(v);
			
			if(type) adj[v].add(u);
			
		}
		
		return adj;
	}
	
	public static void main(String[] args) throws IOException {
		
		Scanner scan=new Scanner(System.in);
		
		int t=scan.nextInt();
		while(t-->0) {
			int n=scan.nextInt();
			int k=scan.nextInt();
			String ans= D(n, k, scan);
		}
		
		
	}
	
	public static void A(long arr[], int n, int k) {
		
		
		int cO=0;
		int cE=0;
		for(long x: arr) {
			if(x%2==0) {
				cE++;
			}else {
				cO++;
			}
		}
		
		if(cO==0) {
			System.out.println("No");
			return;
		}
		
		for(int i=1;i<=Math.min(k,cO);i+=2) {
			
			if(cE>=(k-i)) {
				System.out.println("Yes");
				return;
			}
		}
		
		System.out.println("No");
		
	}
	
	public static long B(String s) {
		
		long c1[]=new long[s.length()];
		long c0[]=new long[s.length()];
		
		if(s.charAt(0)=='1') {
			c1[0]=1;
		}else{
			c0[0]=1;
		}
		
		for(int i=1;i<s.length();i++) {
			
			if(s.charAt(i)=='1') {
				c1[i]=1;
			}else {
				c0[i]=1;
			}
			
			c1[i]+=c1[i-1];
			c0[i]+=c0[i-1];
		}
		
//		System.out.println(Arrays.toString(c0));
//		System.out.println(Arrays.toString(c1));
		
		long min=Integer.MAX_VALUE;
		int n=s.length();
		for(int i=0;i<s.length();i++) {
			
			long ans1= c0[i]+ c1[n-1]-c1[i];
			long ans2= c1[i]+c0[n-1]-c0[i];
			
			min=Math.min(min, Math.min(ans1, ans2));
		}
		
		return min;
		
	}
	
	public static void C(int n, Reader scan, int x) throws IOException {
		
		
		List<Integer> adj[]=new ArrayList[n+1];
		
		for(int i=0;i<n-1;i++) {
			
			int u=scan.nextInt();
			int v=scan.nextInt();
			
			if(adj[u]==null) adj[u]=new ArrayList<>();
			if(adj[v]==null) adj[v]=new ArrayList<>();
			
			adj[u].add(v);
			adj[v].add(u);
		}
		
		if(n==1) {
			System.out.println("Ayush");
			return;
		}
		
		if(adj[x].size()==1||adj[x].size()==0) {
			System.out.println("Ayush");
			return;
		}
		
		int turn=0;
		
		int val= n- 3;
		
		turn= val%2;
		
		if(turn==0) {
			System.out.println("Ashish");
		}else {
			System.out.println("Ayush");
		}
		
	}
	
	public static String D(int n, int k, Scanner scan) {
		
		
		Set<Integer> idx[]=new HashSet[k];
		
		for(int i=0;i<k;i++) {
			idx[i]=new HashSet<>();
			
			int num=scan.nextInt();
			for(int j=0;j<num;j++) {
				idx[i].add(scan.nextInt());
			}
		}
		
		int max=0;
		
		System.out.print("? "+n+" ");
		for(int i=1;i<=n;i++) {
			System.out.print(i+" ");
		}
		System.out.println();
		System.out.println();
		max=scan.nextInt();
		System.out.flush();
		
		
		
		int l=1;
		int h=n;
		
		while(l<h) {
			
			int mid= l+(h-l)/2;
			
			System.out.print("? "+(mid-l+1)+" ");
			for(int i=l;i<=mid;i++) {
				System.out.print(i+" ");
			}
			System.out.println();
			System.out.println();
			int curMax= scan.nextInt();
			System.out.flush();
			
			if(curMax==max) {
				h=mid;
			}else {
				l=mid+1;
			}
		}
		
		
		
		int val=max;
		int ind=0;
		for(;ind<k;ind++) {
			
			if(idx[ind].contains(l)) {
				break;
			}
		}
		
		if(ind==k) {
			System.out.print("! ");
			for(int i=0;i<k;i++) {
				System.out.print(max+" ");
			}
			
			System.out.println();
			System.out.println();
			String s=scan.next();
			System.out.flush();
			
			return s;
		}
		
		
		
		System.out.print("? "+(n-idx[ind].size())+" ");
		for(int j=1;j<=n;j++) {
			if(idx[ind].contains(j)) continue;
			
			System.out.print(j+" ");
		}
		
		System.out.println();
		System.out.println();
		val=scan.nextInt();
		System.out.flush();
		
		
		System.out.print("! ");
		for(int i=0;i<k;i++) {
			if(i==ind) {
				System.out.print(val+" ");
			}else {
				System.out.print(max+" ");
			}
		}
		
		System.out.println();
		System.out.println();
		String s=scan.next();
		System.out.flush();
		
		return s;
		
		
	}

}