#クエリ回数上限+1確認
def floor_log(n):
	a,p=0,1
	while p*2<=n:a,p=a+1,p*2
	return a
O,R=print,range
I=lambda:int(input())
N=I()
q=int(floor_log(N**(N-1))-(N-1)/2+1)+1
n=R(N)
Q=[0]*N
u=set(i+1for i in n)
v=set(u)
def S(Q,k):
	v=set(u)
	for i in n:
		Q[i],k=sorted(v)[k//F[N-i-1]],k%F[N-i-1]
		v.remove(Q[i])
if N<9:
	F=[1]
	for i in n:F+=[F[-1]*(i+1)]
	l,r=0,F[N]
	while l<r-1:
		m,q=(l+r)//2,q-1
		S(Q,m),O("?",*Q)
		if I():l=m
		else:r=m
	S(Q,l)
else:
	for i in R(N-1):
		l,r,X=0,N-i,sorted(v)
		while l<r-1:
			m,j,q=(l+r)//2,i+1,q-1
			t=Q[i]=X[m]
			for k in R(N-i):
				if X[k]!=t:Q[j],j=X[k],j+1
			O("?",*Q)
			t=I()
			if t<1:r=m
			else:l=m
		t=Q[i]=X[l]
		v.remove(t)
	Q[N-1]=v.pop()
while q:
	q-=1
	O("?",*Q)
assert(q==0)
O("!",*Q)