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.io.*;
import java.util.*;
public class Pc{
	public static void main(String[] args) throws IOException{
		FastReader sc = new FastReader();
		BufferedWriter out = new BufferedWriter(new OutputStreamWriter(new FileOutputStream(java.io.FileDescriptor.out),"ASCII"),512);
		int n = sc.nextInt();
		if(n == 1){
			out.write("! 1");
		}else{
			int[] arr = new int[n+1];
			int[] done = new int[n+1];
			int f = 1;
			int s = 2;
			for(int i=0;i<=n;i++){
				out.write("? " + f + " " + s);
				out.write('\n');
				out.flush();
				int a = sc.nextInt();
				out.write("? " + s + " " + f);
				out.write('\n');
				out.flush();
				int b = sc.nextInt();
				if(a>b){
					arr[f] = a;
					done[a] = 1;
					f = s;
					s = f+1;
				}
				else if(b>a){
					arr[s] = b;
					done[b] = 1;
					s++;
				}
				if(s>n) break;
			}
			int left = 0;
			for(int i=1;i<=n;i++){
				if(done[i] == 0) left = i;
			}
			out.write("! ");
			for(int i=1;i<=n;i++){
				if(arr[i] == 0) arr[i] = left;
				out.write(arr[i] + " ");
			}
		}
		out.write('\n');
		out.flush();
	}
}
class FastReader{
	BufferedReader br;
	StringTokenizer st;
	public FastReader(){
		br = new BufferedReader(new InputStreamReader(System.in));
	}
	String next(){
		while(st == null || !st.hasMoreElements()){
			try{ 
				st = new StringTokenizer(br.readLine()); 
			} 
			catch (IOException  e){ 
				e.printStackTrace(); 
			} 
		} 
		return st.nextToken();
	}
	int nextInt(){ 
        return Integer.parseInt(next()); 
	}
    long nextLong(){ 
        return Long.parseLong(next()); 
    }
    double nextDouble(){ 
        return Double.parseDouble(next()); 
    }
    String nextLine(){ 
        String str = ""; 
        try{ 
            str = br.readLine(); 
        } 
		catch (IOException e){ 
            e.printStackTrace(); 
        } 
        return str; 
    }
}