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 EdC {

	public static void main(String[] args) throws Exception{
		int num = 998244353;

		// TODO Auto-generated method stub
 		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
 		PrintWriter out = new PrintWriter(System.out);
 		int n = Integer.parseInt(bf.readLine());
 		int[] ans = new int[n+1];
 		TreeSet<Integer> set = new TreeSet<Integer>();
 		for(int j = 1;j<=n;j++)
 			set.add(j);
 		for(int j = 1;j<n;j++){
 			int i1 = set.pollFirst();
 			int i2 = set.pollFirst();
 			out.println("? " + i1 +" " + i2);
 			out.flush();
 			int a = Integer.parseInt(bf.readLine());
 			out.println("? " + i2 +" " + i1);
 			out.flush();
 			int b = Integer.parseInt(bf.readLine());
 			if (b > a){
 				ans[i2] = b;
 				set.add(i1);
 			}
 			else{
 				ans[i1] = a;
 				set.add(i2);
 			}
 			
 			
 		}
 		out.print("! ");
 		for(int j =1;j<=n;j++){
 			if (ans[j] == 0)
 				out.print(n + " ");
 			else
 				out.print(ans[j]+" ");
 		}
 		out.println();
	 		
 		out.close();
 		
 		
 		
 	}
}
 	
 //hello
//StringJoiner sj = new StringJoiner(" "); 
//sj.add(strings)
//sj.toString() gives string of those stuff w spaces or whatever that sequence is

 		
 		
 		
 		
	


