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.*;
import java.math.*;

public class Solution {

    static BufferedReader in;
    static StringTokenizer stk;
    static int T, N, M, K;
    static PrintWriter out;

    public static void main(String[] args) throws Exception {
        out = new PrintWriter(new OutputStreamWriter(System.out));

        in = new BufferedReader(new InputStreamReader(System.in));
        // Start of User Code

       
        int n = ni();
        int k = ni();
        Map<Integer, Integer> m = new TreeMap<>();
        for(int i = 1; i <= k + 1; i++){
            String s = "?";
            for(int j = 1; j < i; j++){
                s = s + " " + j;
            }
            for(int j = i + 1; j <= k + 1; j++){
                s = s + " " + j;
            }
            out.println(s);
            out.flush();
            int p = ni();
            int q = ni();
            if(!m.containsKey(q)) m.put(q, 0);
            m.put(q, m.get(q) + 1);
        }
        for(Integer key: m.keySet()){
            out.println("! " + (k + 1 - m.get(key)));
            out.flush();
            break;
        }
        
        
        // End of User Code
        out.flush();
        in.close();
    }

    static long modPow(long n, long pow, long mod) {
        return BigInteger.valueOf(n).modPow(BigInteger.valueOf(pow), BigInteger.valueOf(mod)).longValue();
    }

    static long modInv(long n, long mod, boolean isPrimeModuli) {
        if (isPrimeModuli) {
            return modPow(n, mod - 2, mod);
        }
        return BigInteger.valueOf(n).modInverse(BigInteger.valueOf(mod)).longValue();
    }
    // calc factorials
    static long[] fact;

    static void calcFactorials(int n) {
        fact = new long[n + 1];
        fact[0] = 1;
        for (int i = 1; i <= n; i++) {
            fact[i] = fact[i - 1] * i;
        }
    }

    static void calcFactorialsModM(int n, long M) {
        fact = new long[n + 1];
        fact[0] = 1;
        for (int i = 1; i <= n; i++) {
            fact[i] = fact[i - 1] * i;
            fact[i] %= M;
        }
    }

    static long ncr(int n, int r) {
        return fact[n] / (fact[n - r] * fact[r]);
    }

    static long ncrModM(int n, int r, long MOD, boolean isPrimeModuli) {
        return (((fact[n] * modInv(fact[n - r], MOD, isPrimeModuli))%MOD)* modInv(fact[r], MOD, isPrimeModuli))%MOD;
    }

    static long toL(String s) {
        return Long.parseLong(s);
    }

    static long toL(BigInteger n) {
        return n.longValue();
    }

    static int toI(String s) {
        return Integer.parseInt(s);
    }

    static double toD(String s) {
        return Double.parseDouble(s);
    }

    static void printf(String format, Object... args) {
        out.printf(format, args);
    }
    

    static int ni() throws Exception {
        while (stk == null || !stk.hasMoreTokens()) {
            stk = new StringTokenizer(in.readLine());
        }
        return Integer.parseInt(stk.nextToken());
    }

    static long nl() throws Exception {
        while (stk == null || !stk.hasMoreTokens()) {
            stk = new StringTokenizer(in.readLine());
        }
        return Long.parseLong(stk.nextToken());
    }

    static double nd() throws Exception {
        while (stk == null || !stk.hasMoreTokens()) {
            stk = new StringTokenizer(in.readLine());
        }
        return Double.parseDouble(stk.nextToken());
    }

    static String ns() throws Exception {
        while (stk == null || !stk.hasMoreTokens()) {
            stk = new StringTokenizer(in.readLine());
        }
        return stk.nextToken();
    }

    static int min(int a, int b) {
        return a < b ? a : b;
    }

    static int max(int a, int b) {
        return a > b ? a : b;
    }

    static long min(long a, long b) {
        return a < b ? a : b;
    }

    static long max(long a, long b) {
        return a > b ? a : b;
    }

    static int gcd(int a, int b) {
        if (b == 0) {
            return a;
        }
        return gcd(b, a % b);
    }

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