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.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.net.StandardSocketOptions;
import java.security.cert.CollectionCertStoreParameters;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;
import java.util.SortedSet;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;
import java.util.Vector;

import javax.script.ScriptEngine;
import javax.script.ScriptEngineManager;
import javax.script.ScriptException;

public class mai {

	public static int lowerBound(ArrayList<Integer> array, int length, long value) {
		int low = 0;
		int high = length;
		while (low < high) {
			final int mid = (low + high) / 2;
			if (value <= array.get(mid)) {
				high = mid;
			} else {
				low = mid + 1;
			}
		}
		return low;
	}

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			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 long gcd(long a, long b) {
		if (a == 0)
			return b;
		return gcd(b % a, a);
	}

	static long lcm(long a, long b) {
		return (a * b) / gcd(a, b);
	}

	public static void sortbyColumn(int arr[][], int col) {
		// Using built-in sort function Arrays.sort
		Arrays.sort(arr, new Comparator<int[]>() {

			@Override
			// Compare values according to columns
			public int compare(final int[] entry1, final int[] entry2) {

				// To sort in descending order revert
				// the '>' Operator
				// if (entry1[col] > entry2[col])
				// return 1;
				// else //(entry1[col] >= entry2[col])
				// return -1;
				return new Integer(entry1[col]).compareTo(entry2[col]);
				// return entry1[col].
			}
		}); // End of function call sort().
	}

	class pair<X, Y> { // utilizing Java "Generics"
		X _first;
		Y _second;

		public pair(X f, Y s) {
			_first = f;
			_second = s;
		}

		X first() {
			return _first;
		}

		Y second() {
			return _second;
		}

		void setFirst(X f) {
			_first = f;
		}

		void setSecond(Y s) {
			_second = s;
		}
	}

	public static int[] swap(int data[], int left, int right) {
		int temp = data[left];
		data[left] = data[right];
		data[right] = temp;
		return data;
	}

	public static int[] reverse(int data[], int left, int right) {
		while (left < right) {
			int temp = data[left];
			data[left++] = data[right];
			data[right--] = temp;
		}
		return data;
	}

	public static boolean findNextPermutation(int data[]) {
		if (data.length <= 1)
			return false;

		int last = data.length - 2;
		while (last >= 0) {
			if (data[last] < data[last + 1]) {
				break;
			}
			last--;
		}
		if (last < 0)
			return false;

		int nextGreater = data.length - 1;
		for (int i = data.length - 1; i > last; i--) {
			if (data[i] > data[last]) {
				nextGreater = i;
				break;
			}
		}
		data = swap(data, nextGreater, last);
		data = reverse(data, last + 1, data.length - 1);
		return true;
	}

	static long nCr(long n, long r) {
		long x = 1;
		long yu = n - r;
		while (n > r) {
			x = x * n;
			n--;
		}
		while (yu > 0) {
			x /= yu;
			yu--;
		}
		return x;
	}

	/*
	 * static boolean prime[] = new boolean[1000007];
	 * 
	 * public static void sieveOfEratosthenes(int n) {
	 * 
	 * for (int i = 0; i < n; i++) prime[i] = true; * for (int p = 2; p * p <=
	 * n; p++) { // If prime[p] is not changed, then it is a prime if (prime[p]
	 * == true) { // Update all multiples of p for (int i = p * p; i <= n; i +=
	 * p) prime[i] = false; } }
	 * 
	 * // Print all prime numbers // for(int i = 2; i <= n; i++) // { //
	 * if(prime[i] == true) // System.out.print(i + " "); // } }
	 */
	public static final long M = 1000000007;

	public static void main(String[] args) throws NumberFormatException, IOException, ScriptException {
		FastReader sc = new FastReader();
		PrintWriter pr = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		// Scanner scn = new Scanner(System.in);
		//
		int n = sc.nextInt();
		int k = sc.nextInt();
		int arr[] = new int[k + 1];
		HashSet<Integer> ha = new HashSet<>();
		for (int i = 1; i <= (k + 1); i++) {
			pr.flush();
			pr.print("? ");
			for (int j = 1; j <= (k + 1); j++) {
				if (i != j) {
					pr.print(j + " ");
				}
			}
			pr.println();
			pr.flush();
			int a = sc.nextInt();
			arr[i - 1] = sc.nextInt();
			ha.add(arr[i - 1]);
		}
		Object[] ans = ha.toArray();

		int a = 0, b = 0;
		HashMap<Integer, Integer> h = new HashMap<>();
		for (int i = 0; i <= k; i++) {
			if (h.containsKey(arr[i]) == true)
				h.put(arr[i], h.get(arr[i]) + 1);
			else {

				h.put(arr[i], 1);
			}
		}
		//System.out.println(ans[0]+" "+ans[1]);
		//System.out.println(h);
		if ((int) ans[0] > (int) ans[1])
			pr.println("! " +h.get( ans[0]));
		else
			pr.println("! " +h.get( ans[1]));
		//
		// coded to perfection by Rohan Mukhija
		pr.flush();
		pr.close();

	}

	private static boolean possible(long[] arr, int[] f, long mid, long k) {
		long c = 0;

		for (int i = 0; i < arr.length; i++) {
			if ((arr[i] * f[f.length - i - 1]) > mid) {
				c += (arr[i] - (mid / f[f.length - i - 1]));
			}
		}
		// System.out.println(mid+" "+c);
		if (c <= k)
			return true;
		return false;
	}

}