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.*;
public class Solution
{
       public static void main(String []args)
       {
              Scanner sc = new Scanner(System.in);
              int n = sc.nextInt();
              if(n != 1)
              {
                      int arr[] = new int[n+1];
              int maxpos = 1;
              for(int i = 2 ; i <= n ; i++)
              {
                     System.out.println("? " + maxpos + " " + i);
                     System.out.flush();
                     int mod1 = sc.nextInt();
                     System.out.println("? " + i + " " + maxpos);
                     System.out.flush();
                     int mod2 = sc.nextInt();
                     
                     if(mod1 > mod2)
                     {
                            arr[maxpos] = mod1;
                     maxpos = i;
                     }
                     
                     else
                     {
                            arr[i] = mod2;
                     }
              }
              
             arr[maxpos] = n;
              System.out.print("! ");
              for(int i = 1 ; i <= n ; i++)
              {
                  System.out.print(arr[i] + " ");  
              }
             
             
              }
              
              else
              {
                     System.out.println("! 1");
                     System.out.flush();
              }
       }
}