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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class TestTask {
	static long MOD = (long)1e9+7;
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		FastReader sc = new FastReader();
		PrintWriter out = new PrintWriter(System.out);
		
		int test = 1;
		while(test-->0) {
			int n = sc.nextInt();
			int mx = 1;
			int[] p = new int[n+1];
			for(int i = 2;i<=n;i++) {
				out.println("? "+mx+" "+i);
				out.flush();
				int a = sc.nextInt();
				out.println("? "+i+" "+mx);
				out.flush();
				int b = sc.nextInt();
				if(a<b) {
					p[i] = Math.max(a, b);
				} else {
					p[mx] = Math.max(a, b);
					mx = i;
				}
			}
			out.print("! ");
			for(int i = 1;i<=n;i++) {
				out.print((p[i]==0 ? n : p[i])+" ");
			}
		}
		out.close();
	}
	static class FastReader {
		
		StringTokenizer st;
		BufferedReader br;
		
		public FastReader() {
			br = new BufferedReader(
					new InputStreamReader(System.in));
		}
		
		String next() {
			while(st==null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
		
		double nextDouble() { 
			return Double.parseDouble(next()); 
		} 
		
		String nextLine() {
			String s = "";
			while(st==null || st.hasMoreElements()) {
				try {
					s = br.readLine();
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return s;
		}
	}	
}
