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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] ar = new int[n];
        int x = 1;
        for (int i = 1; i < n; i++) {
            System.out.println("? " + x + " " + (i+1));
            System.out.flush();
            int a = sc.nextInt();
            System.out.println("? " + (i+1) + " " + x);
            System.out.flush();
            int b = sc.nextInt();
            if (a>b) {
                ar[x-1] = a;
                x = i+1;
            }
            else {
                ar[i] = b;
            }
        }
        for (int i = 0; i < n; i++) {
            boolean found = false;
            for (int j = 0; j < n; j++) {
                if (ar[j]==i+1) {
                    found=true;
                }
            }
            if (!found) {
                ar[x-1] = i+1;
            }
        }
        System.out.printf("! ");
        for (int i = 0; i < n; i++) {
            System.out.printf(" %d", ar[i]);
        }
    }
    static class Scanner {
        BufferedReader br = new BufferedReader( new InputStreamReader( System.in ) );
        StringTokenizer st = new StringTokenizer( "" );

        public Scanner( InputStream in ) {

        }

        public String next() {
            while ( !st.hasMoreElements() )
                try {
                    st = new StringTokenizer( br.readLine() );
                } catch ( IOException e ) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt( next() );
        }

        public long nextLong() {
            return Long.parseLong( next() );
        }

        public int[] readArray( int n ) {
            int[] a = new int[ n ];
            for ( int i = 0; i < n; i++ ) {
                a[ i ] = nextInt();
            }
            return a;
        }

        public double nextDouble() {
            return Double.parseDouble( next() );
        }

    }
}