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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main
{
    public static void main(String[] args)
    {
        CP sc =new CP();
        int n=sc.nextInt();
        int[] brr =new int[n];
        int first=1;
        boolean[] done =new boolean[n+1];
        for(int i=2;i<=n;i++)
        {
            System.out.println("?"+" "+first+" "+(i));
            int a=sc.nextInt();
            System.out.flush();
            if(a==-1)
            {
                System.exit(0);
            }
            System.out.println("?"+" "+(i)+" "+first);
            int b=sc.nextInt();
            System.out.flush();
            if(b==-1)
            {
                System.exit(0);
            }
            if(a>b)
            {
                brr[first-1]=a;
                done[a]=true;
                for(int j=first+1;j<=n;j++)
                {
                    if(brr[j-1]==0)
                    {
                        first=j;
                        break;
                    }
                }
            }
            else {
                brr[i-1]=b;
                done[b]=true;
            }
        }
        int val=0;
        for(int i=1;i<=n;i++)
        {
            if(!done[i])
                val=i;
        }
        System.out.print("! ");
        for(int i=0;i<n;i++)
        {
            if(brr[i]==0)
                System.out.print(val+" ");
            else
                System.out.print(brr[i]+" ");
        }
        System.out.flush();
    }
    /*****************************************************************************/
    static class CP
    {
        BufferedReader bufferedReader;
        StringTokenizer stringTokenizer;
        public CP() {
            bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        }
        int nextInt() {
            return Integer.parseInt(NNNN());
        }
        long nextLong() {
            return Long.parseLong(NNNN());
        }
        double nextDouble() {
            return Double.parseDouble(NNNN());
        }
        String NNNN() {
            while (stringTokenizer == null || !stringTokenizer.hasMoreElements()) {
                try {
                    stringTokenizer = new StringTokenizer(bufferedReader.readLine());
                }
                catch (IOException  e) {
                    e.printStackTrace();
                }
            }
            return stringTokenizer.nextToken();
        }
        String nextLine() {
            String spl = "";
            try {
                spl = bufferedReader.readLine();
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return spl;
        }
    }
    /*****************************************************************************/
}