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].
#### IMPORTANT LIBRARY ####

############################
### DO NOT USE import random --> 250ms to load the library
############################

### In case of extra libraries: https://github.com/cheran-senthil/PyRival

######################
####### IMPORT #######
######################

from functools import cmp_to_key 
from collections import deque, Counter
from heapq import heappush, heappop
from math import log, ceil

######################
#### STANDARD I/O ####
######################

import sys
import os
from io import BytesIO, IOBase

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
 
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

def print(*args, **kwargs):
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()

def inp():
    return sys.stdin.readline().rstrip("\r\n")  # for fast input
    
def ii():
    return int(inp())

def si():
    return str(inp())

def li(lag = 0):
    l = list(map(int, inp().split()))
    if lag != 0:
        for i in range(len(l)):
            l[i] += lag
    return l
    
def mi(lag = 0):
    matrix = list()
    for i in range(n):
        matrix.append(li(lag))
    return matrix

def lsi(): #string list
    return list(map(str, inp().split()))

def print_list(lista, space = " ", **kwargs):
    print(space.join(map(str, lista)), **kwargs)

######################
### BISECT METHODS ###
######################

def bisect_left(a, x):
    """i tale che a[i] >= x e a[i-1] < x"""
    left = 0
    right = len(a)
    while left < right:
        mid = (left+right)//2
        if a[mid] < x: 
            left = mid+1
        else: 
            right = mid
    return left

def bisect_right(a, x):
    """i tale che a[i] > x e a[i-1] <= x"""
    left = 0
    right = len(a)
    while left < right:
        mid = (left+right)//2
        if a[mid] > x: 
            right = mid
        else: 
            left = mid+1
    return left

def bisect_elements(a, x):
    """elementi pari a x nell'árray sortato"""
    return bisect_right(a, x) - bisect_left(a, x)

######################
### MOD OPERATION ####
######################

MOD = 10**9 + 7
maxN = 5
FACT = [0] * maxN
INV_FACT = [0] * maxN

def add(x, y):
    return (x+y) % MOD
   
def multiply(x, y):
    return (x*y) % MOD

def power(x, y):
    if y == 0:
        return 1
    elif y % 2:
        return multiply(x, power(x, y-1))
    else:
        a = power(x, y//2)
        return multiply(a, a)

def inverse(x):
    return power(x, MOD-2)
   
def divide(x, y):
    return multiply(x, inverse(y))

def allFactorials():
    FACT[0] = 1
    for i in range(1, maxN):
        FACT[i] = multiply(i, FACT[i-1])

def inverseFactorials():
    n = len(INV_FACT)
    INV_FACT[n-1] = inverse(FACT[n-1])
    for i in range(n-2, -1, -1):
        INV_FACT[i] = multiply(INV_FACT[i+1], i+1)

def coeffBinom(n, k):
    if n < k:
        return 0
    return multiply(FACT[n], multiply(INV_FACT[k], INV_FACT[n-k]))

######################
#### GRAPH ALGOS #####
######################

# ZERO BASED GRAPH
def create_graph(n, m, undirected = 1, unweighted = 1):
    graph = [[] for i in range(n)]
    if unweighted:
        for i in range(m):
            [x, y] = li(lag = -1)
            graph[x].append(y)
            if undirected:
                graph[y].append(x)
    else:
        for i in range(m):
            [x, y, w] = li(lag = -1)
            w += 1
            graph[x].append([y,w])
            if undirected:
                graph[y].append([x,w])
    return graph

def create_tree(n, unweighted = 1):
    children = [[] for i in range(n)]
    if unweighted:
        for i in range(n-1):
            [x, y] = li(lag = -1)
            children[x].append(y)
            children[y].append(x)
    else:
        for i in range(n-1):
            [x, y, w] = li(lag = -1)
            w += 1
            children[x].append([y, w])
            children[y].append([x, w])
    return children
   
def dist(tree, n, A, B = -1):
    s = [[A, 0]]
    massimo, massimo_nodo = 0, 0
    distanza = -1
    v = [-1] * n
    while s:
        el, dis = s.pop() 
        if dis > massimo:
            massimo = dis
            massimo_nodo = el
        if el == B:
            distanza = dis
        for child in tree[el]:
            if v[child] == -1:
                v[child] = 1
                s.append([child, dis+1])
    return massimo, massimo_nodo, distanza

def diameter(tree):
    _, foglia, _ = dist(tree, n, 0)
    diam, _, _ = dist(tree, n, foglia)
    return diam
   
def dfs(graph, n, A):
    v = [-1] * n
    s = [[A, 0]]
    v[A] = 0
    while s:
        el, dis = s.pop()
        for child in graph[el]:
            if v[child] == -1:
                v[child] = dis + 1
                s.append([child, dis + 1])
    return v #visited: -1 if not visited, otherwise v[B] is the distance in terms of edges

def bfs(graph, n, A):
    v = [-1] * n
    s = deque()
    s.append([A, 0])
    v[A] = 0
    while s:
        el, dis = s.popleft()
        for child in graph[el]:
            if v[child] == -1:
                v[child] = dis + 1
                s.append([child, dis + 1])
    return v #visited: -1 if not visited, otherwise v[B] is the distance in terms of edges

#FROM A GIVEN ROOT, RECOVER THE STRUCTURE
def parents_children_root_unrooted_tree(tree, n, root = 0):
    q = deque()
    visited = [0] * n
    parent = [-1] * n
    children = [[] for i in range(n)]
    q.append(root)
    while q:
        all_done = 1
        visited[q[0]] = 1
        for child in tree[q[0]]:
            if not visited[child]:
                all_done = 0
                q.appendleft(child)
        if all_done:
            for child in tree[q[0]]:   
                if parent[child] == -1:
                    parent[q[0]] = child
                    children[child].append(q[0])
            q.popleft()
    return parent, children

# CALCULATING LONGEST PATH FOR ALL THE NODES 
def all_longest_path_passing_from_node(parent, children, n):
    q = deque()
    visited = [len(children[i]) for i in range(n)]
    downwards = [[0,0] for i in range(n)]
    upward = [1] * n
    longest_path = [1] * n
    for i in range(n):
        if not visited[i]:
            q.append(i)
            downwards[i] = [1,0]
    while q:
        node = q.popleft()
        if parent[node] != -1:
            visited[parent[node]] -= 1
            if not visited[parent[node]]:
                q.append(parent[node])
        else:
            root = node
        for child in children[node]:   
            downwards[node] = sorted([downwards[node][0], downwards[node][1], downwards[child][0] + 1], reverse = True)[0:2]
    s = [node]
    while s:
        node = s.pop()
        if parent[node] != -1:
            if downwards[parent[node]][0] == downwards[node][0] + 1:
                upward[node] = 1 + max(upward[parent[node]], downwards[parent[node]][1])
            else:
                upward[node] = 1 + max(upward[parent[node]], downwards[parent[node]][0])
        longest_path[node] = downwards[node][0] + downwards[node][1] + upward[node] - min([downwards[node][0], downwards[node][1], upward[node]]) - 1
        for child in children[node]:
            s.append(child)
    return longest_path

### TBD SUCCESSOR GRAPH 7.5
### TBD TREE QUERIES 10.2 da 2 a 4
### TBD ADVANCED TREE 10.3
### TBD GRAPHS AND MATRICES 11.3.3 e 11.4.3 e 11.5.3 (ON GAMES)

######################
## END OF LIBRARIES ##
######################

[n,k] = li()
occ = dict()
for i in range(k+1):
    q = ["?"]
    for j in range(k+1):
        if j != i:
            q.append(j+1)
    print_list(q, flush = True)
    pos, num = li()
    occ[(pos,num)] = occ.get((pos,num),0)+1
ma = 0
freq = 0
for pos, num in occ:
    if num > ma:
        ma = num
        freq = occ[(pos, num)]
print_list(["!", freq], flush = True)
        
    
    

