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.io.*;
import java.util.*;
public class Ishu
{
	static Scanner scan = new Scanner(System.in);
	static void tc()
	{
	int n = scan.nextInt();
	int[] a = new int[n + 1];
	int i = 1,j = 2;
	while(i <= n && j <= n)
		{
		System.out.println("? " + i + " " + j);
		int k1 = scan.nextInt();
		System.out.flush();
		
		System.out.println("? " + j + " " + i);
		int k2 = scan.nextInt();
		System.out.flush();
		
		if(k1 > k2)
			{
			a[i] = k1;
			if(i < j)
				i = j + 1;
			else	
				++i;
			}
		else if(k1 < k2)
			{
			a[j] = k2;
			if(j < i)
				j = i + 1;
			else
				++j;
			}
		}
	int sum = 0;
	for(i=1;i<=n;++i)
		sum += a[i];
	int ele = (n * (n + 1)) / 2 - sum;
	for(i=1;i<=n;++i)
		if(a[i] == 0)
			{
			a[i] = ele;
			break;
			}
	System.out.print("! ");
	for(i=1;i<=n;++i)
		System.out.print(a[i] + " ");
	}
	public static void main(String[] args)
	{
	int t = 1;
	//t = scan.nextInt();
	while(t-- > 0)
		tc();
	}
}