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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  set<long long> s, use;
  for (int i = 1; i < n + 1; i++) s.insert(i), use.insert(i);
  long long out[n + 1];
  while (1) {
    if (s.size() == 1) break;
    auto it = s.begin();
    long long x = *it, xin, yin;
    auto it2 = ++it;
    long long y = *it2;
    cout << "? " << x << " " << y << '\n';
    cout.flush();
    cin >> xin;
    cout << "? " << y << " " << x << '\n';
    cout.flush();
    cin >> yin;
    if (xin > yin) {
      out[x] = xin;
      s.erase(s.find(x));
      use.erase(xin);
    } else {
      out[y] = yin;
      s.erase(s.find(y));
      use.erase(yin);
    }
  }
  out[*s.begin()] = *use.begin();
  cout << "! ";
  for (int i = 1; i < n + 1; i++) cout << out[i] << " ";
  cout << '\n';
  return 0;
}