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

public class C1407 {
	static PrintWriter pw;
	static Scanner sc;

	public static int query(int u, int v) throws IOException {
		pw.println("? "+(u + 1) + " " + (v + 1));
		pw.flush();
		return sc.nextInt();
	}

	public static void main(String[] args) throws IOException {
		sc = new Scanner(System.in);
		pw = new PrintWriter(System.out);
		int n = sc.nextInt();
		Queue<Integer> q = new LinkedList<>();
		for (int i = 0; i < n; i++) {
			q.add(i);
		}
		int[] ans = new int[n];
		boolean[] found = new boolean[n + 1];
		while (q.size() > 1) {
			int x = q.poll();
			int y = q.poll();
			int xmody = query(x, y);
			int ymodx = query(y, x);
			if (xmody < ymodx) {
				ans[y] = ymodx;
				found[ymodx] = true;
				q.add(x);
			} else {
				ans[x] = xmody;
				q.add(y);
				found[xmody] = true;
			}
		}
		int notfound = -1;
		for (int i = 1; i <= n; i++) {
			if (!found[i]) {
				notfound = i;
				break;
			}
		}
		pw.print('!');
		for (int i = 0; i < ans.length; i++) {
			if (ans[i] == 0) {
				pw.print(" " + notfound);
			} else {
				pw.print(" " + ans[i]);
			}
		}
		pw.println();
		pw.flush();
		pw.close();

	}

	static class Scanner {
		BufferedReader br;
		StringTokenizer st;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public Scanner(FileReader f) {
			br = new BufferedReader(f);
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

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

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

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public int[] nextIntArr(int n) throws IOException {
			int[] arr = new int[n];
			for (int i = 0; i < n; i++) {
				arr[i] = Integer.parseInt(next());
			}
			return arr;
		}

	}

}