Problem: 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

Solution (in PYTHON3):

Solution:
import sys

DEBUG = False


def debug(*args):
    if not DEBUG:
        return
    print("\033[0;31m", end="", file=sys.stderr)
    print(*args, file=sys.stderr)
    print("\033[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():
    # Want the nth move to correspond with the nth bit.
    # While reconstructing we just need to know whether to go right or down, so make sure the diagonals alternate bits
    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:
            # last is larger, so a is a[i]
            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()

if __name__ == "__main__":
    solve()