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.*;
import java.util.*;

public class Main {
    static int M = 1_000_000_007;

    private static void testCase(int n, FastScanner in, PrintWriter out) {
        int[] ans = new int[n];
        int isKnown = 0, res1, res2, missing = 0;
        ArrayList<Integer> unknown = new ArrayList<>(n), newUnknown;

        for (int i = 1; i <= n; i++) unknown.add(i);

        while (unknown.size() > 1) {
            newUnknown = new ArrayList<>();

            for (int i = 0; i + 1 < unknown.size(); i += 2) {
                out.println(buildQuery(unknown.get(i), unknown.get(i + 1)));
                out.flush();
                res1 = in.nextInt();

                out.println(buildQuery(unknown.get(i + 1), unknown.get(i)));
                out.flush();
                res2 = in.nextInt();

                //greater value is the element
                if (res1 > res2) {
                    //a[i] = res1
                    ans[unknown.get(i) - 1] = res1;
                    newUnknown.add(unknown.get(i + 1));
                } else { //res1 < res2
                    //a[i + 1] = res2
                    ans[unknown.get(i + 1) - 1] = res2;
                    newUnknown.add(unknown.get(i));
                }
            }

            if (unknown.size() % 2 == 1) newUnknown.add(unknown.get(unknown.size() - 1));

            unknown = newUnknown;
        }

        boolean[] isPresent = new boolean[n + 1];

        for (int an : ans) {
            isPresent[an] = true;
        }

        for (int i = 1; i <= n; i++) {
            if (!isPresent[i]) missing = i;
        }

        for (int i = 0; i < n; i++) {
            if (ans[i] == 0) ans[i] = missing;
        }

        out.println(buildAnswer(ans));
        out.flush();
    }

    private static String buildQuery(int x, int y) {
        return String.format("? %d %d", x, y);
    }

    private static String buildAnswer(int[] ans) {
        StringBuilder sb = new StringBuilder("! ");

        for (int an : ans) {
            sb.append(an);
            sb.append(' ');
        }

        sb.deleteCharAt(sb.length() - 1);
        return sb.toString();
    }

    public static void main(String[] args) {
        FastScanner in = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
//        int t = in.nextInt();  // Scanner has functions to read ints, longs, strings, chars, etc.
        //in.nextLine();
//        for (int tt = 1; tt <= t; ++tt) {
            int n = in.nextInt();

            testCase(n, in, out);

            out.println();
//        }

        out.close();
    }

    private 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();
        }

        char[] readCharArray(int n) {
            char[] arr = new char[n];
            try {
                br.read(arr);
                br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return arr;
        }

        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[] readLongArray(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++) a[i] = nextLong();
            return a;
        }

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