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.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Scanner;
import java.util.stream.Collectors;

public class Bunny {
    public static void main(String args[]) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int[] data = new int[n];

        if (n == 1) {
            data[0] = 1;
        } else {
            int l = 0;
            int r = 1;

            while (l < n && r < n) {
                String req = "? " + (l + 1) + " " + (r + 1);
//                System.out.printf("? %d %d\n", l + 1, r + 1);
                System.out.println(req);
                System.out.flush();
                int a = in.nextInt();

                req = "? " + (r + 1) + " " + (l + 1);
//                System.out.printf("? %d %d\n", r + 1, l + 1);
                System.out.println(req);
                System.out.flush();
                int b = in.nextInt();

                if (a > b) {
                    data[l] = a;
                    l = Math.max(l, r) + 1;
                } else {
                    data[r] = b;
                    r = Math.max(l, r) + 1;
                }
            }

            data[Math.min(l, r)] = n;
        }

        print(data);
    }

    private static void print(int[] data) {
        String result = "! " + Arrays.stream(data).mapToObj(String::valueOf).collect(Collectors.joining(" "));
        System.out.println(result);
    }
}
