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:
/*
Keep solving problems.
*/

import java.util.*;
import java.io.*;

public class Interactive {

    BufferedReader br;
    PrintWriter out;
    StringTokenizer st;
    boolean eof;
    final long MOD = 1000L * 1000L * 1000L + 7;
    int[] dx = {0, -1, 0, 1};
    int[] dy = {1, 0, -1, 0};

    void solve() throws IOException {
        int n = nextInt();
        if (n == 1) {
            outln("! " + 1);
            return;
        } else if (n == 2) {
            outln("? 1 2");
            int k = nextInt();
            if (k == 1) {
                outln("! " + 1 + " " + 2);
            } else {
                outln("! " + 2 + " " + 1);
            }
            return;
        }

//        TreeSet<Integer> ts = new TreeSet<>();
//        for (int i = 0; i < n; i++) {
//            ts.add(i + 1);
//        }

        int first = 1;
        int[] res = new int[n];
        for (int i = 1; i <= n - 1; i++) {
            int second = i + 1;
            print(first, second);
            int k1 = nextInt();
            print(second, first);
            int k2 = nextInt();

            if (k1 < k2) {
                res[second - 1] = k2;
            } else {
                res[first - 1] = k1;
                first = second;
            }
        }

        int sum = 0;
        for (int i = 1; i <= n; i++) {
            sum += i;
        }

        for (int v : res) {
            sum -= v;
        }
        for (int i = 0; i < n; i++) {
            if (res[i] == 0) {
                res[i] = sum;
            }
        }

        out("! ");
        for (int i = 0; i < n - 1; i++) {
            out(res[i] + " ");
        }
        outln(res[n - 1]);
    }

    void print(int x, int y) {
        outln("? " + x + " " + y);
    }

    void shuffle(int[] a) {
        int n = a.length;
        for(int i = 0; i < n; i++) {
            int r = i + (int) (Math.random() * (n - i));
            int tmp = a[i];
            a[i] = a[r];
            a[r] = tmp;
        }
    }

    long gcd(long a, long b) {
        while(a != 0 && b != 0) {
            long c = b;
            b = a % b;
            a = c;
        }
        return a + b;
    }

    private void outln(Object o) {
        out.println(o);
        out.flush();
    }
    private void out(Object o) {
        out.print(o);
        out.flush();
    }
    public Interactive() throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        out.close();
    }
    public static void main(String[] args) throws IOException {
        new Interactive();
    }

    public long[] nextLongArr(int n) throws IOException{
        long[] res = new long[n];
        for(int i = 0; i < n; i++)
            res[i] = nextLong();
        return res;
    }
    public int[] nextIntArr(int n) throws IOException {
        int[] res = new int[n];
        for(int i = 0; i < n; i++)
            res[i] = nextInt();
        return res;
    }
    public String nextToken() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                eof = true;
                return null;
            }
        }
        return st.nextToken();
    }
    public String nextString() {
        try {
            return br.readLine();
        } catch (IOException e) {
            eof = true;
            return null;
        }
    }
    public int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }
    public long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }
    public double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }
}