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.*;
import java.io.*;
public class Choclate {

	static BufferedReader br;
	static StringTokenizer tokenizer;
//	static ArrayList<Integer> tester = new ArrayList<Integer>();
	public static void main(String[] args) throws Exception {
		br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);
		int n = nextInt();
//		for(int i = 1; i < n + 1; i++)
//			tester.add(i);
//		Collections.shuffle(tester);
		Stack<Integer> arr = new Stack<Integer>();
		for(int i = 0; i < n; i++)
			arr.push(i + 1);
		int[] ans = new int[n];
		while(arr.size() > 1) {
			int i = arr.pop();
			int j = arr.pop();
			out.println("? " + i + " " + j);
			out.flush();
			int t = nextInt();
//			int t = tester.get(i - 1) % tester.get(j - 1);
			out.println("? " + j + " " + i);
			out.flush();
			int t1 = nextInt();
//			int t1 = tester.get(j - 1) % tester.get(i - 1);
			if(t1 > t) {
				ans[j - 1] = t1;
				arr.push(i);
			} else {
				ans[i - 1] = t;
				arr.push(j);
			}
		}
//		System.out.println(Arrays.toString(ans));
		boolean[] has = new boolean[n];
		for(int i = 0; i < n; i++) {
			if(ans[i] != 0)
				has[ans[i] - 1] = true;
		}
		for(int i = 0; i < n; i++)
			if(!has[i])
				ans[arr.pop() - 1] = i + 1;
		out.print("! ");
		for(int i = 0; i < n; i++)
			out.print(ans[i] + " ");
		out.println();
		out.flush();
		out.close();
	}
	public static String next() throws IOException {
		while (tokenizer == null || !tokenizer.hasMoreTokens()) {
			String line = br.readLine();
			if (line == null)
				throw new IOException();
			tokenizer = new StringTokenizer(line);
		}
		return tokenizer.nextToken();
	}

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

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

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