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.Scanner;

public class Test
{
	public static void main(String[] args)
	{
		Scanner scanner = new Scanner(System.in);
		int n = scanner.nextInt();
		int[] a = new int[n];

		if(n == 1)
		{
			System.out.println("! 1");
			System.out.flush();
			return;
		}

		int i=0, j=1;

		for(int w=0; w<2*n; w++)
		{
			System.out.println("? " + (i+1) + " " + (j+1));
			System.out.flush();
			int xi = scanner.nextInt();

			System.out.println("? " + (j+1) + " " + (i+1));
			System.out.flush();
			int xj = scanner.nextInt();

			if(xi > xj)
			{
				a[i] = xi;
				while(i<n && a[i] != 0) i++;
				if(i == j) i++;
			}
			else
			{
				a[j] = xj;
				while(j<n && a[j] != 0) j++;
				if(i == j) j++;
			}

			if(i==j || i==n || j==n) break;
		}
		StringBuilder sb = new StringBuilder();
		sb.append("! ");

		for(int w=0; w<n; w++)
		{
			if(a[w] == 0)
			{
				a[w] = n;
			}
			sb.append(a[w]).append(" ");
		}

		System.out.println(sb.toString().trim());
		System.out.flush();
	}

}