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.*;
import java.io.*;

public class C{
    static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter pw=new PrintWriter(System.out);

    public static void main(String[] args) throws Exception{

        int N=Integer.parseInt(br.readLine());
        if(N==1){
           pw.print("! ");
           pw.println(1+" ");
           pw.flush();
           pw.close();
            return;
        }
        int[] arr=new int[N];
        boolean[] elements=new boolean[N+1];//1 indexed for simplicity

        int calls=0,prev_ptr=0,next_ptr=1,pmodn,nmodp;
        while(calls<2*N){
            pw.println( "? " + (prev_ptr+1) + " " + (next_ptr+1) );
            pw.flush();
            pmodn=Integer.parseInt(br.readLine());
            pw.println( "? " + (next_ptr+1) + " " + (prev_ptr+1) );
            pw.flush();
            nmodp=Integer.parseInt(br.readLine());
            calls+=2;
            if( pmodn == 0 && nmodp == 1 ){
                elements[1]=true;
                arr[next_ptr]=1;
                prev_ptr+=2;
                next_ptr=prev_ptr+1;
            }else if( pmodn == 1 && nmodp == 0 ){
                elements[1]=true;
                arr[prev_ptr]=1;
                prev_ptr+=2;
                next_ptr+=2;
            }else{
                if(pmodn>nmodp){
                    elements[pmodn]=true;
                    arr[prev_ptr]=pmodn;
                }else{
                    elements[nmodp]=true;
                    arr[next_ptr]=nmodp;
                }
                prev_ptr+=2;
                next_ptr+=2;
            }
            if(prev_ptr>=N || next_ptr>=N)break;// Need to do an array iteration now
        }

        
        while( calls < (2*N) ){
            prev_ptr=-1;next_ptr=-1;
            for(int i=0;i<N;i++){
                if(arr[i]==0 && prev_ptr==-1)
                    prev_ptr=i;
                else if( arr[i]==0 && next_ptr==-1)
                    next_ptr=i;
            }

            if(prev_ptr !=-1 && next_ptr !=-1 ){
                pw.println( "? " + (prev_ptr+1) + " " + (next_ptr+1) );
                pw.flush();
                pmodn=Integer.parseInt(br.readLine());
                pw.println( "? " + (next_ptr+1) + " " + (prev_ptr+1) );
                pw.flush();
                nmodp=Integer.parseInt(br.readLine());
                calls+=2;
                if(pmodn>nmodp){
                    elements[pmodn]=true;
                    arr[prev_ptr]=pmodn;
                }else{
                    elements[nmodp]=true;
                    arr[next_ptr]=nmodp;
                }
            }else break;
        }

        // 1 element can still remain missing... We need no do any API call for that and prev ptr would already be pointing to that

        int i;
        for(i=0;i<N;i++){
            if(arr[i]==0)break;
        }

        for(int j=1;j<=N;j++){
            if(elements[j]==false){
                arr[i]=j;
                break;
            }
        }

        

        pw.print("! ");
        for(int k:arr)
            pw.print(k+" ");
        pw.flush();
        pw.close();
    }
}