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.awt.List;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.PriorityQueue;
import java.util.Random;
import java.util.Scanner;

public class C {

	public static void main(String[] srgs) throws NumberFormatException, IOException {
		Scanner nik = new Scanner(System.in);

		int n = nik.nextInt();

		int[] a = new int[n];

		int l = 1;
		int r = 2;
		long n1 = (n * (n + 1)) / 2;
		while (r <= n) {

			System.out.println("? " + l + " " + r);
			System.out.flush();
			int g1 = nik.nextInt();

			System.out.println("? " + r + " " + l);
			System.out.flush();
			int g2 = nik.nextInt();

			if (g1 > g2) {
				a[l - 1] = g1;
				n1 -= g1;
				l = r;
			} else {
				a[r - 1] = g2;
				n1 -= g2;
			}
			r++;
		}

		StringBuilder st = new StringBuilder();
		st.append("! ");
		for (int i = 0; i < n; i++) {
			if (a[i] == 0) {
				st.append(n1 + " ");
			} else {
				st.append(a[i] + " ");
			}
		}
		System.out.println(st);
		System.out.flush();

	}
}