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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, i, j;
  scanf("%d", &n);
  bool vis[10001];
  memset(vis, false, sizeof vis);
  vector<int> listFalse;
  for (i = 1; i <= n; i++) {
    listFalse.push_back(i);
  }
  if (n == 1) {
    printf("! 1\n");
    fflush(stdout);
    return 0;
  }
  bool visBil[10001];
  memset(visBil, false, sizeof visBil);
  int a[10001];
  while (listFalse.size() >= 2) {
    int idx1 = listFalse.back();
    listFalse.pop_back();
    int idx2 = listFalse.back();
    listFalse.pop_back();
    printf("? %d %d\n", idx1, idx2);
    fflush(stdout);
    int ans1, ans2;
    scanf("%d", &ans1);
    printf("? %d %d\n", idx2, idx1);
    fflush(stdout);
    scanf("%d", &ans2);
    if (ans1 < ans2) {
      a[idx2] = ans2;
      listFalse.push_back(idx1);
      vis[idx2] = true;
      visBil[ans2] = true;
    } else {
      a[idx1] = ans1;
      listFalse.push_back(idx2);
      vis[idx1] = true;
      visBil[ans1] = true;
    }
  }
  int miss = 0;
  for (i = 1; i <= n; i++) {
    if (!visBil[i]) {
      miss = i;
    }
  }
  for (i = 1; i <= n; i++) {
    if (!vis[i]) {
      a[i] = miss;
    }
  }
  printf("!");
  for (i = 1; i <= n; i++) {
    printf(" %d", a[i]);
  }
  printf("\n");
  fflush(stdout);
  return 0;
};