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 C {
    public static void main(String[] args) throws IOException{
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        int n = Integer.parseInt(f.readLine());
        int[] arr = new int[n];
        int ptr1 = 0;
        int ptr2 = 1;
        int[] perm = new int[n];
        if(n == 1){
            out.println("! 1");
            out.flush();
            out.close();
            return;
        }
        for(int i = 0; i < n-1; i++){
            out.println("? " + (ptr1+1) + " " + (ptr2+1));
            out.flush();
            int ans1 = Integer.parseInt(f.readLine());
            out.println("? " + (ptr2+1) + " " + (ptr1+1));
            out.flush();
            int ans2 = Integer.parseInt(f.readLine());
            if(ans1 > ans2){
                perm[ptr1] = ans1;
                ptr1++;
                while(ptr1 < n && perm[ptr1] != 0) ptr1++;
                if(ptr1 == ptr2) ptr2++;
            }else{
                perm[ptr2] = ans2;
                ptr2++;
                while(ptr2 < n && perm[ptr2] != 0) ptr2++;
                if(ptr1 == ptr2) ptr1++;
            }
        }
        StringBuilder ans = new StringBuilder();
        for(int i = 0; i < n; i++){
            if(perm[i] == 0) ans.append(n+" ");
            else ans.append(perm[i]+" ");
        }
        out.println("! " + ans.toString().trim());
        out.close();
    }
}
