Problem: This is an interactive problem.

Ayush devised a new scheme to set the password of his lock. The lock has k slots where each slot can hold integers from 1 to n. The password P is a sequence of k integers each in the range [1, n], i-th element of which goes into the i-th slot of the lock.

To set the password of his lock, Ayush comes up with an array A of n integers each in the range [1, n] (not necessarily distinct). He then picks k non-empty mutually disjoint subsets of indices S_1, S_2, ..., S_k (S_i \underset{i ≠ j} ∩ S_j = ∅) and sets his password as P_i = max_{j ∉ S_i} A[j]. In other words, the i-th integer in the password is equal to the maximum over all elements of A whose indices do not belong to S_i.

You are given the subsets of indices chosen by Ayush. You need to guess the password. To make a query, you can choose a non-empty subset of indices of the array and ask the maximum of all elements of the array with index in this subset. You can ask no more than 12 queries.

Input

The first line of the input contains 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 contains two integers n and k (2 ≤ n ≤ 1000, 1 ≤ k ≤ n) — the size of the array and the number of subsets. k lines follow. The i-th line contains an integer c (1 ≤ c < n) — the size of subset S_i, followed by c distinct integers in the range [1, n] — indices from the subset S_i.

It is guaranteed that the intersection of any two subsets is empty.

Interaction

To ask a query print a single line: 

  * In the beginning print "? c " (without quotes) where c (1 ≤ c ≤ n) denotes the size of the subset of indices being queried, followed by c distinct space-separated integers in the range [1, n]. 



For each query, you will receive an integer x — the maximum of value in the array among all the indices queried. If the subset of indices queried is invalid or you exceeded the number of queries (for example one of the indices is greater than n) then you will get x = -1. In this case, you should terminate the program immediately.

When you have guessed the password, print a single line "! " (without quotes), followed by k space-separated integers — the password sequence.

Guessing the password does not count towards the number of queries asked.

After this, you should read a string. If you guess the password correctly, you will receive the string "Correct". In this case, you should continue solving the remaining test cases. If the guessed password is incorrect, you will receive the string "Incorrect". In this case, you should terminate the program immediately.

The interactor is not adaptive. The array A does not change with queries.

After printing a query do not forget to output end of 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 documentation for other languages.



Hacks

To hack the solution use the following test format:

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

The first line of each test case should contain two integers n and k (2 ≤ n ≤ 1000, 1 ≤ k ≤ n) — the size of the array and the number of subsets. The next line should consist of n space separated integers in the range [1, n] — the array A. k lines should follow. The i-th line should contain an integer c (1 ≤ c < n) — the size of subset S_i, followed by c distinct integers in the range [1, n] — indices from the subset S_i.

The intersection of any two subsets has to be empty.

Example

Input


1
4 2
2 1 3
2 2 4

1

2

3

4

Correct

Output


? 1 1

? 1 2

? 1 3

? 1 4

! 4 3

Note

The array A in the example is [1, 2, 3, 4]. The length of the password is 2. The first element of the password is the maximum of A[2], A[4] (since the first subset contains indices 1 and 3, we take maximum over remaining indices). The second element of the password is the maximum of A[1], A[3] (since the second subset contains indices 2, 4).

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

Solution (in JAVA):

Solution:
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStreamWriter;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.PriorityQueue;
import java.util.Map.Entry;

import java.util.Random;
import java.util.Scanner;
import java.util.TreeSet;

public final class CF_646_D2_D2
{


	static boolean verb=true;
	static void log(Object X){if (verb) System.err.println(X);}
	static void log(Object[] X){if (verb) {for (Object U:X) System.err.print(U+" ");System.err.println("");}}
	static void log(int[] X){if (verb) {for (int U:X) System.err.print(U+" ");System.err.println("");}}
	static void logWln(int[] X){if (verb) {for (int U:X) System.err.print(U+" ");}}
	static void log(int[] X,int L){if (verb) {for (int i=0;i<L;i++) System.err.print(X[i]+" ");System.err.println("");}}
	static void log(long[] X){if (verb) {for (long U:X) System.err.print(U+" ");System.err.println("");}}

	static void logWln(Object X){if (verb) System.err.print(X);}
	static void info(Object o){	System.out.println(o);}
	static void output(Object o){outputWln(""+o+"\n");	}
	static void outputFlush(Object o){try {out.write(""+ o+"\n");out.flush();} catch (Exception e) {}}
	static void outputWln(Object o){try {out.write(""+ o);} catch (Exception e) {}}


	static void logBin(int[] tm) {for (int x:tm) logWln(bin(8,x)+" ");log("");}

	static String bin(int L,int x) {
		String s=Integer.toBinaryString(x);
		while (s.length()<L) s="0"+s;
		return s;
	}

	static long powerMod(long b,long e,long m){
		long x=1;
		while (e>0) {
			if (e%2==1)
				x=(b*x)%m;
			b=(b*b)%m;
			e=e/2;
		}
		return x; 
	}


	static class Composite implements Comparable<Composite>{

		int val;
		int idx;


		public int compareTo(Composite X) {
			if (val!=X.val)
				return val-X.val;
			return idx-X.idx;
		}


		public Composite(int val, int idx) {
			this.val = val;
			this.idx = idx;
		}







	}

	
	static void solve(int n,int l,int r,int[][] S,int[] rem,int bigMax) {
		//log("solve rec");
		if (l==r) {
			// nothing to do
			// take the complimentary
		
			System.out.print("? "+(n-S[l].length)+" ");
			for (int i=0;i<S.length;i++) {
				if (i!=l)
					for (int x:S[i])
						System.out.print((x+1)+" ");
			}
			for (int x:rem)
				System.out.print((x+1)+" ");
			System.out.println();
			System.out.flush();
			int altMax=sc.nextInt();
			System.out.print("! ");
			for (int i=0;i<S.length;i++) {
				if (i==l)
					System.out.print(altMax+" ");
				else
					System.out.print(Math.max(bigMax,altMax)+" ");
			}
			System.out.println();
			System.out.flush();
			String verdict=sc.next();
			
			return;
		}
		int mid=(l+r)/2;
		int sz=0;
		for (int i=0;i<=mid;i++) {
			sz+=S[i].length;
		}
		System.out.print("? "+sz+" ");
		for (int i=0;i<=mid;i++) {
			for (int x:S[i])
				System.out.print((x+1)+" ");
		}
		System.out.println();
		System.out.flush();
		int quer=sc.nextInt();
		if (quer==bigMax)
			solve(n,l,mid,S,rem,bigMax);
		else
			solve(n,mid+1,r,S,rem,bigMax);
	}
	static void solve(int n,int[][] S,int[] rem) {
		// is the max in rem ?
		// if so we are done
		// otherwise, more action
		int k=S.length;
		log("query global");
		System.out.print("? "+n+" ");
		for (int i=0;i<n;i++)
			System.out.print((i+1)+" ");
		System.out.println();
		System.out.flush();
		int bigMax=sc.nextInt();

		// now iterate
		solve(n,0,k-1,S,rem,bigMax);
	}
	
	// Global vars
	static BufferedWriter out;
	static InputReader reader;

	static int[][] dd={{-1,0},{0,1},{1,0},{0,-1}};

	static long MX=1000000001;

	static long mod=1000000007;

	static Scanner sc;
	
	static void process() throws Exception {

		//out = new BufferedWriter(new OutputStreamWriter(System.out));
		//reader = new InputReader(System.in);
		sc=new Scanner(System.in);

		int T=sc.nextInt();
		for (int t=0;t<T;t++) {
			int n=sc.nextInt();
			int k=sc.nextInt();
			int[][] S=new int[k][];
			int[] mem=new int[n];
			int tot=0;
			for (int i=0;i<k;i++) {
				int c=sc.nextInt();
				log("c:"+c);
				tot+=c;
				S[i]=new int[c];
				for (int j=0;j<c;j++) {
					S[i][j]=sc.nextInt()-1;
					mem[S[i][j]]=1;
				}
			
			}
			
			int[] rem=new int[n-tot];
			//log("rem.size:"+rem.length);
			int st=0;
			for (int u=0;u<n;u++)
				if (mem[u]==0)
					rem[st++]=u;
			solve(n,S,rem);
			
			
		}

		/*
		try {
			out.close();
		} catch (Exception Ex) {
		}
		*/

	}




	public static void main(String[] args) throws Exception {
		process();

	}

	static final class InputReader {
		private final InputStream stream;
		private final byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;

		public InputReader(InputStream stream) {
			this.stream = stream;
		}

		private int read() throws IOException {
			if (curChar >= numChars) {
				curChar = 0;
				numChars = stream.read(buf);
				if (numChars <= 0) {
					return -1;
				}
			}
			return buf[curChar++];
		}

		public final String readString() throws IOException {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			StringBuilder res = new StringBuilder();
			do {
				res.append((char) c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}

		public final String readString(int L) throws IOException {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			StringBuilder res = new StringBuilder(L);
			do {
				res.append((char) c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}

		public final int readInt() throws IOException {
			int c = read();
			boolean neg = false;
			while (isSpaceChar(c)) {
				c = read();
			}
			char d = (char) c;
			// log("d:"+d);
			if (d == '-') {
				neg = true;
				c = read();
			}
			int res = 0;
			do {
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			// log("res:"+res);
			if (neg)
				return -res;
			return res;

		}

		public final long readLong() throws IOException {
			int c = read();
			boolean neg = false;
			while (isSpaceChar(c)) {
				c = read();
			}
			char d = (char) c;
			// log("d:"+d);
			if (d == '-') {
				neg = true;
				c = read();
			}
			long res = 0;
			do {
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			// log("res:"+res);
			if (neg)
				return -res;
			return res;

		}

		private boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}
	}

}