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.*;
import java.lang.*;
import java.io.*;
import java.math.*;
public final class CF
{
    public static void main(String[]args)throws IOException
    {
        FastReader ob=new FastReader();
        //int t=ob.nextInt();
        StringBuffer sb=new StringBuffer();
        //while(t-->0)
        {
            int n=ob.nextInt();
            int a[]=new int[n];
            Queue<Integer> q=new LinkedList<>();
            for(int i=1;i<=n;i++)
            q.add(i);
            while(q.size()>1)
            {
               int i=q.poll();
               int j=q.poll();
               System.out.println("? "+i+" "+j);
               int x=ob.nextInt();
               System.out.println("? "+j+" "+i);
               int y=ob.nextInt();
               int p=-1,s;
               if(x<y)
               {
                    p=j;
                    s=i;
               }
               else
               {
                    p=i;
                    s=j;
               }
               a[p-1]=Math.max(x, y);
               q.add(s);
            }
            int x=q.poll();
            a[x-1]=n;
            System.out.print("! ");
            for (int i = 0; i < a.length; i++) {
                 System.out.print(a[i]+" ");
            }
        }
        System.out.println(sb);
    }
}
class FastReader {
    BufferedReader br;
    StringTokenizer st;
    public FastReader() {
        br = new BufferedReader(new InputStreamReader(System.in));
    }

    public String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e)  {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }
    public String nextLine()
    {
        String s="";
        try {
        s=br.readLine();
        } catch (IOException e)  {
                e.printStackTrace();
            }
        return s;
    }
    public int nextInt() {
        return Integer.parseInt(next());
    }
    public long nextLong() {
        return Long.parseLong(next());
    }
    public double nextDouble() {
        return Double.parseDouble(next());
    }
}