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 C {
 
  public static void main(String[] args) { 
        solve();
      out.close();
      System.out.flush();
  }

  private static void solve() {
      int n = sc.nextInt();
      int[] res = new int[n + 1];
      Queue<Integer> queue = new LinkedList<>();
      int sum = 0;
      for(int i = 1; i <= n; i++) {
        sum += i;
        queue.offer(i);
      }

      while(queue.size() >= 2) {
          int i = queue.poll();
          int j = queue.poll();
          System.out.println("? " + i + " " + j);
          System.out.println();

          System.out.flush();
          int a = sc.nextInt();

          System.out.println("? " + j + " " + i);
          System.out.println();

          System.out.flush();
          int b = sc.nextInt();

          if(a < b) {
            res[j] = b;
            queue.offer(i);
            sum -= b;
          } else {
            res[i] = a;
            sum -= a;
            queue.offer(j);
          }
          
          System.out.flush();
      }

      if(queue.size() >= 1) {
        res[queue.poll()] = sum;
      }
      StringBuilder sb = new StringBuilder();
      for(int i = 1; i <= n; i++)
        sb.append(res[i] + " ");
      System.out.println("! " + sb.toString());
  }

  public static void sort(int[] a) {
    int l = 0, r = a.length;
      ArrayList<Integer> list=new ArrayList<>();
      for (int i=l ; i<r; i++) list.add(a[i]);
      Collections.sort(list);
      for (int i=l; i<r; i++) a[i]=list.get(i-l);
  }
  
  public static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
 
  public static FastScanner sc = new FastScanner();

  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());
      }
  }
}