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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main
{
    public static void main(String[] args)
    {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskC
    {
        public void solve(int testNumber, Scanner in, PrintWriter out)
        {
            int T = 1;
            while (T-- > 0)
            {
                solveOne(in, out);
            }
        }

        private void solveOne(Scanner in, PrintWriter out)
        {
            int n = in.nextInt();
            int[] nums = new int[n + 1];
            int largestId = 1;
            for (int i = 2; i <= n; i++)
            {
                out.println("? " + largestId + " " + i);
                out.flush();
                int mod1 = in.nextInt();
                out.println("? " + i + " " + largestId);
                out.flush();
                int mod2 = in.nextInt();
                if (mod1 > mod2)
                {
                    nums[largestId] = mod1;
                    largestId = i;
                } else
                {
                    nums[i] = mod2;
                }
            }
            nums[largestId] = n;
            StringBuilder ans = new StringBuilder("!");
            for (int i = 1; i < nums.length; i++)
            {
                ans.append(" " + nums[i]);
            }
            out.println(ans);
            out.flush();
        }

    }
}

