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
//package round669;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Scanner;

public class C {
	Scanner in;
	PrintWriter out;
	String INPUT = "";
	
	void solve()
	{
		int n = ni();
		int x = 1;
		int[] ret = new int[n];
		long s = 0;
		for(int y = 2;y <= n;y++) {
			out.println("? " + x + " " + y);
			out.flush();
			int xy = ni();
			out.println("? " + y + " " + x);
			out.flush();
			int yx = ni();
			if(xy > yx) {
				ret[x-1] = xy;
				s += xy;
				x = y;
			}else {
				ret[y-1] = yx;
				s += yx;
			}
		}
		ret[x-1] = (int)(n*(n+1)/2 - s);
		out.print("!");
		for(int v : ret) {
			out.print(" " + v);
		}
		out.println();
		out.flush();
	}
	
	void run() throws Exception
	{
		in = oj ? new Scanner(System.in) : new Scanner(INPUT);
		out = new PrintWriter(System.out);

		long s = System.currentTimeMillis();
		solve();
		out.flush();
		tr(System.currentTimeMillis()-s+"ms");
	}
	
	public static void main(String[] args) throws Exception
	{
		new C().run();
	}
	
	int ni() { return Integer.parseInt(in.next()); }
	long nl() { return Long.parseLong(in.next()); }
	double nd() { return Double.parseDouble(in.next()); }
	boolean oj = System.getProperty("ONLINE_JUDGE") != null;
	void tr(Object... o) { if(!oj)System.out.println(Arrays.deepToString(o)); }
}
