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.util.*;
import java.io.*;
 
public class D {
 
	static final boolean stdin = true;
	static final String filename = "";
	static FastScanner br;
	static PrintWriter pw;
 
	public static void main(String[] args) throws IOException {
 
		if (stdin) {
			br = new FastScanner();
			pw = new PrintWriter(new OutputStreamWriter(System.out));
		} else {
			br = new FastScanner(filename + ".in");
			pw = new PrintWriter(new FileWriter(filename + ".out"));
		}
 
		Solver solver = new Solver();
		solver.solve(br, pw);
		pw.close();
	}
 
	static class Solver {
		static long mod = (long) (1e9);
 
		public void solve(FastScanner br, PrintWriter pw) throws IOException {
			int n = br.ni();
			int k = br.ni();
			if (k == 1) {
				System.out.println("! " + 1);
				System.exit(0);
			}
			TreeMap<Point, Integer> h = new TreeMap<>();
			int z = k + 1;
			int x = 1;
			int buffer = 0;
			while (z-- > 0) {
				int[] q = new int[k];
				for (int i = 1; i <= k + 1; i++) {
					if (i != x) {
						q[i - 1 - buffer] = i;
					} else {
						buffer = 1;
					}
				}
				System.out.print("? ");
				for (int i : q) {
					System.out.print(i + " ");
				}
				System.out.println();
				System.out.flush();
				int ind = br.ni();
				int val = br.ni();
				if (h.containsKey(new Point(ind, val))) {
					h.put(new Point(ind, val), h.get(new Point(ind, val)) + 1);
				} else {
					h.put(new Point(ind, val), 1);
				}
				x++;
				buffer = 0;
			}
			int s = -1;
			Point w = null;
			for (Point i : h.keySet()) {
				if (h.get(i) > s) {
					s = h.get(i);
					w = i;
				}
			}
			Point r = null;
			for (Point i : h.keySet()) {
				if (i != w) {
					r = i;
				}
			}
			if (w.val > r.val) {
				System.out.println("! " + s);
			} else {
				System.out.println("! " + (k + 1 - s));
			}
			System.out.flush();
		}
 
		static long gcd(long a, long b) {
			if (a > b)
				return gcd(b, a);
			if (a == 0)
				return b;
			return gcd(b % a, a);
		}
 
		static long lcm(long a, long b) {
			return a * (b / gcd(a, b));
		}
 
		static long pow(long a, long b) {
			if (b == 0)
				return 1L;
			long val = pow(a, b / 2);
			if (b % 2 == 0)
				return val * val % mod;
			else
				return val * val % mod * a % mod;
		}
 
	}
 
	static class Point implements Comparable<Point> {
		int ind;
		int val;
 
		Point(int a, int b) {
			this.ind = a;
			this.val = b;
		}
 
		@Override
		public int compareTo(Point o) {
			return this.val - o.val;
		}
 
		public boolean equals(Object obj) {
			if (obj instanceof Point) {
				Point other = (Point) obj;
				return ind == other.ind && val == other.val;
			}
			return false;
		}
 
		public int hashCode() {
			return 65536 * ind + val + 4733 * 0;
		}
 
	}
 
	public static class FastScanner {
		BufferedReader br;
		StringTokenizer st;
 
		public FastScanner(String s) {
			try {
				br = new BufferedReader(new FileReader(s));
			} catch (FileNotFoundException e) {
				e.printStackTrace();
			}
		}
 
		public FastScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
 
		ArrayList<Integer>[] ng(int n, int e) {
			ArrayList<Integer>[] adj = new ArrayList[n];
			for (int i = 0; i < n; i++) {
				adj[i] = new ArrayList<Integer>();
			}
			for (int i = 0; i < e; i++) {
				int a = ni() - 1;
				int b = ni() - 1;
				adj[a].add(b);
				adj[b].add(a);
			}
			return adj;
		}
 
		Integer[] nIa(int n) {
			Integer[] arr = new Integer[n];
			for (int i = 0; i < n; i++) {
				arr[i] = ni();
			}
			return arr;
		}
 
		int[] nia(int n) {
			int[] arr = new int[n];
			for (int i = 0; i < n; i++) {
				arr[i] = ni();
			}
			return arr;
		}
 
		Long[] nLa(int n) {
			Long[] arr = new Long[n];
			for (int i = 0; i < n; i++) {
				arr[i] = nl();
			}
			return arr;
		}
 
		long[] nla(int n) {
			long[] arr = new long[n];
			for (int i = 0; i < n; i++) {
				arr[i] = nl();
			}
			return arr;
		}
 
		String[] nsa(int n) {
			String[] arr = new String[n];
			for (int i = 0; i < n; i++) {
				arr[i] = nt();
			}
			return arr;
		}
 
		String nt() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
 
		int ni() {
			return Integer.parseInt(nt());
		}
 
		long nl() {
			return Long.parseLong(nt());
		}
 
		double nd() {
			return Double.parseDouble(nt());
		}
 
	}
}