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.Arrays;
import java.util.Scanner;
public class T3 {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
//        int n = s.nextInt();
        for (int i=0;i<1;i++) {
            mainPrint(s);
        }
    }
    public static void mainPrint(Scanner s){
        int n=s.nextInt();
        if(n==1) {
            System.out.println("! 1");
            return;
        }

        int[] ans = new int[n];
        int p1=0,p2=1;
        for (int i=0;i<n-1;i++){
            query(p1+1,p2+1);
            int get1 = s.nextInt();
            query(p2+1,p1+1);
            int get2 = s.nextInt();
            if(get1<get2) ans[p2] = Math.max(get1,get2);
            else ans[p1] = Math.max(get1,get2);
            while (p1<n && (ans[p1]!=0 || p1==p2)) p1++;
            while (p2<n && (ans[p2]!=0 || p1==p2)) p2++;
        }
        if(p1<n) ans[p1]=n;
        else ans[p2]=n;
        StringBuilder s1 = new StringBuilder();
        s1.append("!");
        for (int i=0;i<n;i++) {
            s1.append(" ");
            s1.append(ans[i]);
        }
        System.out.println(s1.toString());

    }
    public static void query(int i,int j){
        System.out.println("? "+i+" "+j);
        System.out.flush();
    }
}
