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.io.*;
import java.util.*;

public class test{
    public static void main(String args[]){
       FastScanner fs = new FastScanner();
      
           int n=fs.nextInt();
           LinkedList<Integer> st=new LinkedList<>();
           for(int i=1;i<=n;i++){
               st.add(i);
           }
           int []arr=new int[n];
           while(st.size()>1){
               int i=st.poll();
               int j=st.poll();
               System.out.println("? " + i +" " + j);
               System.out.flush();
               int mod1=fs.nextInt();
               System.out.println("? " + j +" " + i);
               System.out.flush();
               int mod2=fs.nextInt();
               if(mod1>mod2){
                   arr[i-1]=mod1;
                   st.add(j);
               }else{
                   arr[j-1]=mod2;
                   st.add(i);
               }
           }
           arr[st.poll()-1]=n;
           System.out.print("!");
           for(int i=0;i<n;i++){
               System.out.print(" "+arr[i]);
           }
    }

    // static int lower_bound(int arr[],int key){ // first index not smaller than key
    //     int n=arr.length;
    //     int lo=0,hi=n-1;
    //     int mid;
    //     if(arr[n-1]<key){
    //         return n;
    //     }
    //     while(lo<hi){
    //         mid=lo+ (hi-lo)/2;
    //         if(arr[mid]<key){
    //             lo=mid+1;
    //         }else{
    //             hi=mid;
    //         }
    //     }
    //     return lo;        
    // }


    static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
    }
    
}