Problem: 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

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class C {
    public static void main(String[] args) {
        FastInput input = new FastInput();
        PrintWriter output = new PrintWriter(System.out);
        int n = input.nextInt();
        int greater = 1;
        int[] array= new int[n + 1];
        for (int i = 2; i <= n; i++) {
            int mod = ask(greater, i, input, output);
            int mod2 = ask(i, greater, input, output);
            if (mod > mod2) {
                array[greater] = mod;
                greater = i;
            } else {
                array[i] = mod2;
            }
        }
        array[greater] = n;
        /*
        for (int i = 1; i <= n; i++) {
            if (array[i] == 0) {
                int mod = ask(i, greater, input, output);
                array[i] = mod;
            }
        }
        
         */
        output.print("! ");
        for (int i = 1; i <= n; i++) output.print(array[i] + " ");
        output.println();
        output.flush();
    }

    static int ask(int n1, int n2, FastInput input, PrintWriter output) {
        output.println("? " + n1 + " " + n2);
        output.flush();
        return input.nextInt();
    }

    static class FastInput {
        BufferedReader br;
        StringTokenizer st;

        public FastInput() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

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

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

    }
}