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.*;
import java.util.*;

public class ChocolateBunny {

	public static void main(String[] args) {
        FastScanner in = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
        int n = in.nextInt();
        int p[] = new int[n];
        int prev = 1;
        for(int i=2;i<=n;i++) {
        	out.println("? "+prev+" "+i); out.flush();
        	int x = in.nextInt();
        	out.println("? "+i+" "+prev); out.flush();
        	int y = in.nextInt();
        	if(x>y) {
        		p[prev-1] = x; 
        		prev = i;
        	}
        	else
        		p[i-1] = y;
        }
        out.print("! "); out.flush();
        for(int i=0;i<n;i++) {
        	if(p[i]==0) out.print(n+" ");
        	else out.print(p[i]+ " ");
        }
        out.flush();
	}
	
	static class FastScanner {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		
		String next() {
			while(!st.hasMoreTokens())
				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());
		}
		
		int[] readArray(int n) {
			int a[] = new int[n];
			for(int i=0;i<n;i++) a[i] = nextInt();
			return a;
		}
		
		Integer[] readArray(long n) {
		    Integer a[] = new Integer[(int)n];
			for(int i=0;i<n;i++) a[i] = nextInt();
			return a;
		}
		
		long[] readArray(int n,long x) {
			long a[] = new long[n];
			for(int i=0;i<n;i++) a[i] = nextLong();
			return a;
		}
		
	}
	
	static class Pair implements Comparable<Pair> {
		int x, y;
		Pair(int a, int b){ x = a; y = b; }
		int getKey() { return x; }
		int getVal() { return y; }
		@Override
		public int compareTo(Pair o) {
			if(o.getVal() - this.getVal()>0)
				return -1;
			else if(o.getVal() - this.getVal()<0)
				return 1;
			else 
				return 0;
		}
		
	}
	
	static boolean arrayEquals(char a[], char b[]) {
		int n = a.length;
		boolean verdict = true;
		for(int i=0;i<n;i++) {
			if(a[i]!=b[i]) {
				verdict = false;break;
			}
		}
		return verdict;
	}
	
	static long lcm(long a, long b) {
		return (a*b)/gcd(a,b);
	}
	
	static long gcd(long a, long b) {
		if(b==0) return a;
		else return gcd(b,a%b);
	}
	
	static long hashInt(int x,int y) {
		return x*(1_000_000_000L)+y;
	}
}
