Problem: 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

Solution (in JAVA):

Solution:
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
// import java.math.BigInteger;
import java.util.*;

public class Solution {
    public static void main(String[] args) {
        int T = 1;
    TEST:
        while (T-- > 0) {
            int n = in.nextInt();
            int[] perm = new int[n];
            int i = 0, j = n-1, sum = n*(n+1)/2;
            while (i < j) {
                int x = query(i+1, j+1), y = query(j+1, i+1);
                if (x > y) {
                    perm[i++] = x;
                    sum -= x;
                } else {
                    perm[j--] = y;
                    sum -= y;
                }
            }
            if (i == j) {
                perm[i] = sum;
            }
            out.print("! ");
            for (int p : perm) {
                out.print(p + " ");
            }
            out.println();
        }
        out.flush();
    }
    
    static int query(int x, int y) {
        out.println("? " + x + " " + y);
        out.flush();
        return in.nextInt();
    }

    // Handle I/O
    static FastScanner in = new FastScanner();
    static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;
        public FastScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
            st = null;
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        int[] readIntArray(int size) {
            int[] arr = new int[size];
            for (int i = 0; i < size; i++) {
                arr[i] = nextInt();
            }
            return arr;
        }

        long nextLong() {
            return Long.parseLong(next());
        }
    }
}