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.*;
import java.io.*;

public class C {
	public static void main(String[] args) throws Exception {
		int n = i();
		ArrayList<Integer> list = new ArrayList<>();
		for (int i = 1; i <= n; i++) {
			list.add(i);
		}
		int[] ans = new int[n + 1];
		while (list.size() >= 2) {
			ArrayList<Integer> list2 = new ArrayList<>();
			for (int i = 0; i + 1 < list.size(); i += 2) {
				int a = list.get(i), b = list.get(i + 1);
				int q1 = query(a, b), q2 = query(b, a);
				if (q1 < q2) {
					ans[b] = q2;
					list2.add(a);
				} else {
					ans[a] = q1;
					list2.add(b);
				}
			}
			if (list.size() % 2 == 1) {
				list2.add(list.get(list.size() - 1));
			}
			list = list2;
		}

		if (list.size() == 1) {
			int sum = 0;
			for (int i = 1; i <= n; i++) {
				sum += ans[i];
			}
			int x = n * (n + 1) / 2 - sum;
			ans[list.get(0)] = x;
		}

		out.print("!");
		for (int i = 1; i <= n; i++) {
			out.print(" " + ans[i]);
		}
		out.println();

		finish();
	}

	static int query(int i, int j) throws Exception {
		out.println("? " + i + " " + j);
		out.flush();
		return i();
	}

	static BufferedReader in;
	static StringTokenizer st = new StringTokenizer("");
	static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
	static {
		// try {
		// 	in = new BufferedReader(new FileReader("test.in"));
		// } catch (Exception e) {
			in = new BufferedReader(new InputStreamReader(System.in));
		// }
	}
	static void check() throws Exception {
		while (!st.hasMoreTokens()) st = new StringTokenizer(in.readLine());
	}
	static int i() throws Exception {check(); return Integer.parseInt(st.nextToken());}
	static String s() throws Exception {check(); return st.nextToken();}
	static double d() throws Exception {check(); return Double.parseDouble(st.nextToken());}
	static long l() throws Exception {check(); return Long.parseLong(st.nextToken());}
	static int[] a(int n) throws Exception {check(); int[] b = new int[n]; for (int i = 0; i < n; i++) b[i] = i(); return b;}
	static void finish(boolean b, Object o) {if (b) finish(o);}
	static void finish(Object o) {out.println(o); finish();}
	static void finish() {out.close(); System.exit(0);}
}
