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.io.*;
public class C1407{
    public static void main(String []args) throws Exception {
        
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb = new StringBuilder();
        int n = PI(br.readLine());
        int arr[] = new int[n];
        Arrays.fill(arr,-1);
        int curr=0;
        int next = n-1;
        int c = 1;
        int visited[] = new int[n+1];
        while((c++)<n){
            String query = "? "+(1+curr)+" "+(1+next);
            System.out.println(query);
            System.out.flush();
            int val1 = PI(br.readLine());
            String query2 = "? "+(1+next)+" "+(1+curr);
            System.out.println(query2);
            System.out.flush();
            int val2 = PI(br.readLine());
            if(val1>val2){
                arr[curr] = val1;
                curr++;
                visited[val1]++;
            }
            if(val1<val2){
                arr[next] = val2;
                next--;
                visited[val2]++;
            }
            
        }
        for(int i=1;i<=n;i++){
            if(visited[i]==0)
                arr[curr] = i;
        }
        for(int i=0;i<n;i++)
            sb.append(arr[i]+" ");
        System.out.println("! "+sb.toString());

    }
    private static int PI(String s){
        return Integer.parseInt(s);
    }
    private static int[] PA(String s){
        String temp[] = s.split(" ");
        int arr[] = new int[temp.length];
        for(int i=0;i<temp.length;i++){
            arr[i] = PI(temp[i]);
        }
        return arr;
    }
}