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.LinkedList;
import java.util.List;
import java.util.Scanner;

public class CodeForces2010 {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		algo(sc);
		sc.close();
	}
	
	public static void algo(Scanner sc) {
		int n = sc.nextInt();
		if(n==1) {
			System.out.println("! 1");
			return;
		}
		
			
		List<Integer> r = new LinkedList<>();
		for(int i = 0; i<n; i++)
			r.add(-1);
		int i = 0; int j = 1; int k = 0;
		while(k<2*n && j<n) {
			System.out.println("? "+ (i+1) + " " + (j+1) + "\n");
			System.out.flush();
			int a = sc.nextInt();
			k++;
			System.out.println("? "+ (j+1) + " " + (i+1) + "\n");
			System.out.flush();
			int b = sc.nextInt();
			k++;
			if(a==-1 || b==-1) {
				System.exit(1);
			}else if(a>b) {
				r.set(i, a);
				i = j;
				j++;
			} else if(b>a) {
				r.set(j, b);
				j++;
			}
		}
		r.set(r.indexOf(-1), n);
		StringBuilder res = new StringBuilder("");
		for(Integer c: r)
			res.append(c + " ");
		System.out.println("! " + res.toString());
	}	
}
