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.lang.*;
import java.io.*;

public class Main
{
	PrintWriter out = new PrintWriter(System.out);
	BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer tok = new StringTokenizer("");
    String next() throws IOException {
        if (!tok.hasMoreTokens()) { tok = new StringTokenizer(in.readLine()); }
        return tok.nextToken();
    }
    int ni() throws IOException { return Integer.parseInt(next()); }
    long nl() throws IOException { return Long.parseLong(next()); }
    
    void solve() throws IOException {
        int n=ni();
        int[]A=new int[n+1];
        boolean[]F=new boolean[n+1];
        
        int p=1;
        int q=2;
        for (int i=1;i<n;i++) {
            while (A[p]>0 || p==q) p++;
            while (A[q]>0 || q==p) q++;
            
            int a=query(p,q);
            int b=query(q,p);
            
            if (a>b) { A[p]=a; F[a]=true; }
            else { A[q]=b; F[b]=true; }
            
        }
        
        int last=0;
        for (int i=1;i<=n;i++)
            if (!F[i]) last=i;
            
        out.print("! ");
        for (int i=1;i<=n;i++) {
            if (A[i]>0) out.print(A[i]+" ");
            else out.print(last+" ");
        }
        out.println();
        out.flush();
    }
    
    int query(int p,int q) throws IOException {
        out.println("? "+p+" "+q);
        out.flush();
        int r=ni();
        if (r==-1) System.exit(-1);
        return r;
    }
    
    public static void main(String[] args) throws IOException {
        new Main().solve();
    }
}