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
#include <bits/stdc++.h>
using namespace std;
long long fastpow(long long x, long long y) {
  if (y == 0) return 1;
  long long a = fastpow(x, y / 2);
  if (y & 1) {
    return (x * (a * a));
  }
  return (a * a);
}
void solve() {
  int n;
  cin >> n;
  int t = n;
  int ans[n + 1];
  memset(ans, 0, sizeof(ans));
  set<int> notfound;
  unordered_map<int, bool> present;
  for (int i = 1; i <= n; i++) {
    notfound.insert(i);
  }
  while (notfound.size() >= 2) {
    int a = *notfound.begin();
    notfound.erase(a);
    int b = *notfound.begin();
    notfound.erase(b);
    int ans1, ans2;
    cout << "? " << a << " " << b << endl;
    cin >> ans1;
    cout << "? " << b << " " << a << endl;
    cin >> ans2;
    if (ans1 > ans2) {
      ans[a] = ans1;
      present[ans[a]] = true;
      notfound.insert(b);
    } else {
      ans[b] = ans2;
      present[ans[b]] = true;
      notfound.insert(a);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (present[i] == false) {
      ans[*notfound.begin()] = i;
      break;
    }
  }
  cout << "! ";
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long q;
  q = 1;
  while (q--) {
    solve();
  }
  return 0;
}
