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.ArrayList;
import java.util.Scanner;

public class Main {
	
	public static void main(String[] args)
	{
		Scanner s=new Scanner(System.in);
		
		int n=s.nextInt();
		
		if(n==1)
		{
			System.out.println("! 1");
		}
		else
		{
			int[] arr=new int[n+1];
	 		
			int max=0;
			
			System.out.println("? 1 2");
			int a=s.nextInt();
			
			System.out.println("? 2 1");
			int b=s.nextInt();
			
			if(a>b)
			{
				arr[1]=a;
				max=2;
			}
			else
			{
				arr[2]=b;
				max=1;
			}
			
			for(int i=3;i<=n;i++)
			{
				System.out.println("? "+max+" "+i);
				int aa=s.nextInt();
				System.out.println("? "+i+" "+max);
				int bb=s.nextInt();
				
				if(aa>bb)
				{
					arr[max]=aa;
					max=i;
				}
				else
				{
					arr[i]=bb;
				}
			}
			
			arr[max]=n;
			
			System.out.print("! ");
			
			for(int i=1;i<n+1;i++)
			{
				System.out.print(arr[i]+" ");
			}
		}
		
		System.out.flush();
		
	}
	
}