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.*;
import java.util.stream.IntStream;
public class ChocolateBunny{
	static BufferedWriter out=new BufferedWriter(new OutputStreamWriter(System.out));
//	static FastReader in=new FastReader();
	static Scanner in=new Scanner(System.in);
	public static void main(String[] args) throws IOException{
		int t;
//		t=in.nextInt();
		t=1;
		while(t-->0) solve();
	}
	static int askQuery(int i, int j){
		System.out.println("? " + (i+1) + " " + (j+1));
		System.out.flush();
		return in.nextInt();
	}
	static void solve() throws IOException{
		int n=in.nextInt();
		int[] res=new int[n];
		int ref=0;
		for(int i=1; i<n; ++i){
			int res1=askQuery(ref, i);
			int res2=askQuery(i, ref);
			if(res1>res2){
				res[ref]=res1;
				ref=i;
			}
			else{
				res[i]=res2;
			}
		}
		res[ref]=n;
		out.write("! ");
		for(int x: res){
			out.write(x +" ");
		}
		out.write("\n");
		out.flush();
		out.close();
	}
	static 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;
		}
	}
}

