Problem: This problem is interactive.

We have hidden an array a of n pairwise different numbers (this means that no two numbers are equal). You can get some information about this array using a new device you just ordered on Amazon. 

This device can answer queries of the following form: in response to the positions of k different elements of the array, it will return the position and value of the m-th among them in the ascending order.

Unfortunately, the instruction for the device was lost during delivery. However, you remember k, but don't remember m. Your task is to find m using queries to this device. 

You can ask not more than n queries.

Note that the array a and number m are fixed before the start of the interaction and don't depend on your queries. In other words, interactor is not adaptive.

Note that you don't have to minimize the number of queries, and you don't need to guess array a. You just have to guess m.

Input

The first line contains two integers n and k (1≤ k < n ≤ 500) — the length of the array and the number of the elements in the query.

It is guaranteed that number m satisfies 1≤ m ≤ k, elements a_1, a_2, ..., a_n of the array satisfy 0≤ a_i ≤ 10^9, and all of them are different.

Interaction

You begin the interaction by reading n and k.

To ask a question about elements on positions x_1, x_2, ..., x_k, in a separate line output

? x_1 x_2 x_3 ... x_k

Numbers in the query have to satisfy 1 ≤ x_i ≤ n, and all x_i have to be different. Don't forget to 'flush', to get the answer.

In response, you will receive two integers pos and a_{pos} — the position in the array a of the m-th in ascending order element among a_{x_1}, a_{x_2}, ..., a_{x_k}, and the element on this position.

In case your query is invalid or you asked more than n queries, the program will print -1 and will finish interaction. You will receive a Wrong answer verdict. Make sure to exit immediately to avoid getting other verdicts.

When you determine m, output 

! m

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.



Hack format

For the hacks use the following format:

The first line has to contain three integers n, k, m (1 ≤ m ≤ k < n ≤ 500) — the length of the array, number of elements in the query, and which in the ascending order number the device returns.

In the next line output n integers a_1, a_2, ..., a_n (0≤ a_i ≤ 10^9) — the elements of the array. They have to be pairwise different.

Example

Input


4 3
4 9
4 9
4 9
1 2

Output


? 2 3 4
? 1 3 4
? 1 2 4
? 1 2 3
! 3

Note

In the example, n = 4, k = 3, m = 3, a = [2, 0, 1, 9].

Solution (in JAVA):

Solution:

import java.io.OutputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.*;
import java.io.BufferedReader;
import java.io.InputStreamReader;
//import java.math.*;
 
/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
 
public class Main {
	
	static int MD= 1000000007 ;
    public static void main(String[] args) throws IOException {
    	OutputStream outputStream = System.out;
        InputReader in = new InputReader();
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }
   
    static class TaskC {
   
	    public void solve(int testNumber, InputReader in, PrintWriter out) throws IOException {
	    	int t=1;
	    	while(t-->0) { 
	    		int n=in.nextInt();
	    		int k=in.nextInt();
	    		int low=Integer.MAX_VALUE;
	    		int res=0;
	    		for(int i=0;i<=k;i++) {
	    			System.out.print("? ");
	    			
	    			int[] a=new int[k];
	    			a[k-1]=-1;
	    			int cnt=0;
	    			for(int j=0;j<k;) {
	    				if(i!=cnt) {
	    					a[j]=cnt;
	    					j++;
	    				}
	    				cnt++;
	    			}
	    			if(a[k-1]==-1) {
	    				a[k-1]=k;
	    			}
	    			for(int j=0;j<k;j++) {
	    				System.out.print((a[j]+1)+" ");
	    			}
	    			System.out.println();
	    			System.out.flush();
	    			int midx=in.nextInt();
	    			int min=in.nextInt();
	    			
	    			if(min<low) {
	    				low=min;
	    				res=1;
	    			}
	    			else if(min==low) {
	    				res++;
	    			}
	    		}
	    		System.out.println("! "+(k+1-res));
	    				
	    		
	    		
	    		
	    		
	    		
	        }
	 
	    	
	    }
	   
    
    }
   
   
 
 
    
    static class InputReader
    {
        BufferedReader br;
        StringTokenizer st;
 
        public InputReader()
        {
            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;
        }
    
    }
    static void sort(int[] a, int low, int high) 
    {
        int N = high - low;         
        if (N <= 1) 
            return; 
        int mid = low + N/2; 
        sort(a, low, mid); 
        sort(a, mid, high); 
        int[] temp = new int[N];
        int i = low, j = mid;
        for (int k = 0; k < N; k++) 
        {
            if (i == mid)  
                temp[k] = a[j++];
            else if (j == high) 
                temp[k] = a[i++];
            else if (a[j]<a[i]) 
                temp[k] = a[j++];
            else 
                temp[k] = a[i++];
        }    
        for (int k = 0; k < N; k++) 
            a[low + k] = temp[k];         
    }
    static void lsort(long[] a, int low, int high) 
    {
        int N = high - low;         
        if (N <= 1) 
            return; 
        int mid = low + N/2; 
        lsort(a, low, mid); 
        lsort(a, mid, high); 
        long[] temp = new long[N];
        int i = low, j = mid;
        for (int k = 0; k < N; k++) 
        {
            if (i == mid)  
                temp[k] = a[j++];
            else if (j == high) 
                temp[k] = a[i++];
            else if (a[j]<a[i]) 
                temp[k] = a[j++];
            else 
                temp[k] = a[i++];
        }    
        for (int k = 0; k < N; k++) 
            a[low + k] = temp[k];         
    }
    public static long fexp(long x, long y, long p) {
        long res = 1;      
        
       
        x = x % p;  
      
        while (y > 0) 
        { 
         
            if((y & 1)==1) 
                res = (res * x) % p; 
      
          
            y = y >> 1;  
            x = (x * x) % p;  
        } 
        return res; 
        
    }
    static int gcd(int a, int b) 
    { 
        if (b == 0) 
            return a; 
        return gcd(b, a % b);  
          
    }
    static long lgcd(long a, long b) 
    { 
        if (b == 0) 
            return a; 
        return lgcd(b, a % b);  
          
    }
  
}