This is an interactive problem.

We hid from you a permutation p of length n, consisting of the elements from 1 to n. You want to guess it. To do that, you can give us 2 different indices i and j, and we will reply with p_{i} mod p_{j} (remainder of division p_{i} by p_{j}).

We have enough patience to answer at most 2 ⋅ n queries, so you should fit in this constraint. Can you do it?

As a reminder, a permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).

Input

The only line of the input contains a single integer n (1 ≤ n ≤ 10^4) — length of the permutation.

Interaction

The interaction starts with reading n. 

Then you are allowed to make at most 2 ⋅ n queries in the following way: 

  * "? x y" (1 ≤ x, y ≤ n, x ≠ y). 



After each one, you should read an integer k, that equals p_x mod p_y. 

When you have guessed the permutation, print a single line "! " (without quotes), followed by array p and quit.

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.



Exit immediately after receiving "-1" and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.

Hack format

In the first line output n (1 ≤ n ≤ 10^4). In the second line print the permutation of n integers p_1, p_2, …, p_n.

Example

Input


3

1

2

1

0

Output


? 1 2

? 3 2

? 1 3

? 2 1

! 1 3 2

Source: codeforces | Difficulty: MEDIUM_HARD
Tags: Amortized analysis

Mathematics

Constructive algorithms


Solutions:
import sys
DEBUG = False

def debug(*args):
	if not DEBUG:
		return
	print('\x1b[0;31m', end='', file=sys.stderr)
	print(*args, file=sys.stderr)
	print('\x1b[0m', end='', file=sys.stderr)
	sys.stderr.flush()

def readInt():
	line = input()
	while line == '':
		line = input()
	result = int(line)
	return result
cache = {}

def query(i, j):
	if (i, j) not in cache:
		print('? ' + str(i + 1) + ' ' + str(j + 1), file=sys.stdout)
		sys.stdout.flush()
		if not DEBUG:
			x = readInt()
			debug('query', i, j, ':', x)
		else:
			x = REAL[i] % REAL[j]
			debug('query', i, j, '\t', REAL[i], '%', REAL[j], ':', x)
		cache[i, j] = x
	return cache[i, j]

def answer(arr):
	print('! ' + ' '.join((str(x) for x in arr)), file=sys.stdout)
	sys.stdout.flush()
	debug('ans', arr)

def solve():
	if DEBUG:
		cache.clear()
		N = len(REAL)
		debug('Testing', N, REAL)
	else:
		N = readInt()
	if N == 1:
		answer([1])
		exit()
	ans = [-1 for i in range(N)]
	last = 0
	for i in range(1, N):
		a = query(i, last)
		b = query(last, i)
		if a > b:
			ans[i] = a
			if DEBUG:
				assert REAL[last] > REAL[i]
		else:
			ans[last] = b
			if DEBUG:
				assert REAL[last] < REAL[i]
			last = i
	for i in range(N):
		if ans[i] == -1:
			ans[i] = N
	answer(ans)
	assert len(cache) <= 2 * N
	return ans
if DEBUG:
	import random
	random.seed(0)
	for _ in range(1000):
		N = 5
		REAL = list(range(1, N + 1))
		random.shuffle(REAL)
		assert solve() == REAL
	exit()
solve()