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.io.PrintWriter;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        
        var sc = new Scanner(System.in);
        var pw = new PrintWriter(System.out);
        
        int n = Integer.parseInt(sc.next());
        
        var ans = new int[n+1];
        var used = new boolean[n+1];
        int x = 1;
        int y;
        for(int i = 2; i <= n; i++){
            y = i;
            
            pw.println("? " + x + " " + y);
            pw.flush();
            int a = Integer.parseInt(sc.next());
            if(a == -1) return;
            
            pw.println("? " + y + " " + x);
            pw.flush();
            int b = Integer.parseInt(sc.next());
            if(b == -1) return;
            
            if(a > b){
                ans[x] = a;
                used[a] = true;
                x = y;
            }else{
                ans[y] = b;
                used[b] = true;
            }
        }
        pw.print("!");
        for(int i = 1; i <= n; i++){
            if(i != x){
                pw.print(" " + ans[i]);
            }else{
                for(int j = 1; j <= n; j++){
                    if(!used[j]){
                        pw.print(" " + j);
                        break;
                    }
                }
            }
        }
        pw.println();
        pw.flush();
    }
}