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
import java.util.Arrays;
import java.util.Scanner;

public class C_ChocolateBunny {
    final static int NOT_SET = -777;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] perm = new int[n + 1];
        Arrays.fill(perm, NOT_SET);

        int todo = n;
        while (todo > 1) {
            int first = NOT_SET;
            for (int i = 1; i <= n; i++) {
                if (perm[i] == NOT_SET) {
                    if (first == NOT_SET) first = i;
                    else {
                        System.out.println("? " + first + " " + i);
                        int response1 = sc.nextInt();
                        System.out.println("? " + i + " " + first);
                        int response2 = sc.nextInt();

                        if (response1 < response2) {
                            perm[i] = response2;
                        } else {
                            perm[first] = response1;
                        }
                        todo -= 1;

                        first = NOT_SET;
                    }
                }
            }
        }

        StringBuilder sb = new StringBuilder("! ");
        for (int i = 1; i <= n; i++) {
            sb.append(perm[i] == NOT_SET ? n : perm[i]).append(" ");
        }
        System.out.println(sb);
    }
}
