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;
void solve() {
  int n;
  cin >> n;
  set<int> quer;
  for (int i = 1; i <= n; i++) {
    quer.insert(i);
  }
  int x, y;
  int ans[n + 1];
  for (int i = 0; i <= n; i++) ans[i] = -1;
  while (quer.size() > 1) {
    x = *quer.begin();
    int fir, sec;
    quer.erase(x);
    if (quer.size() == 0) break;
    while (quer.size() >= 1) {
      int u = *quer.begin();
      if (u != x) {
        cout << "? " << x << " " << u << endl;
        cin >> fir;
        if (fir == -1) exit(0);
        cout << "? " << u << " " << x << endl;
        cin >> sec;
        if (sec == -1) exit(0);
        if (fir > sec) {
          ans[x] = fir;
          break;
        } else {
          ans[u] = sec;
          quer.erase(u);
        }
      }
      if (quer.size() == 0) break;
    }
  }
  cout << "! ";
  for (int i = 1; i <= n; i++) {
    if (ans[i] == -1)
      cout << n << " ";
    else
      cout << ans[i] << " ";
  }
  cout.flush();
  exit(0);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}