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 JAVA):

Solution:
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int mod = 1000000007;

        int arr[] = new int[n];

        Queue<Integer> queue = new LinkedList<>();
        HashSet<Integer> set = new HashSet<>();

        for (int i = 0; i < n; i++) {
            queue.offer(i);
            set.add(i + 1);
        }

        while (queue.size() > 1) {
            int idx1 = queue.poll();
            int idx2 = queue.poll();
            int one = -1;
            int two = -1;
            System.out.print("? " + (idx1 + 1) + " " + (idx2 + 1) + "\n");
            System.out.flush();
            one = sc.nextInt();
            System.out.print("? " + (idx2 + 1) + " " + (idx1 + 1) + "\n");
            System.out.flush();
            two = sc.nextInt();
            if (one > two) {
                arr[idx1] = one;
                queue.offer(idx2);
                set.remove(one);
            } else {
                arr[idx2] = two;
                queue.offer(idx1);
                set.remove(two);
            }
        }

        int last = queue.poll();
        for (int i = 0; i < n; i++) {
            if (set.contains(i + 1)){
                arr[last] = i + 1;
                break;
            }
        }
        StringBuilder sb = new StringBuilder();
        sb.append("!");
        for (int i = 0; i < n; i++) {
            sb.append(" " + (arr[i]));
        }
        System.out.print(sb.toString());
        System.out.flush();
    }
}