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;
const long long N = 1e5 + 2, mod = 1e9 + 7, inf = 1e9 + 10;
long long a[N];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  if (n == 1) {
    cout << "! 1";
    return 0;
  }
  long long mx = -inf, pos, lastpos;
  for (long long i = 1; i <= n - 1; i += 2) {
    long long res, res1;
    cout << "? " << i << " " << i + 1 << endl;
    cin >> res;
    cout << "? " << i + 1 << " " << i << endl;
    cin >> res1;
    if (res > res1)
      a[i] = res;
    else
      a[i + 1] = res1;
    if (a[i] >= (n + 1) / 2) pos = i;
    if (a[i + 1] >= (n + 1) / 2) pos = i + 1;
    if (a[i] == 0) lastpos = i;
    if (a[i + 1] == 0) lastpos = i + 1;
  }
  if (n % 2) {
    long long res, res1;
    cout << "? " << lastpos << " " << n << endl;
    cin >> res;
    cout << "? " << n << " " << lastpos << endl;
    cin >> res1;
    if (res > res1)
      a[lastpos] = res;
    else
      a[n] = res1;
    if (a[lastpos] >= (n + 1) / 2) pos = lastpos;
    if (a[n] >= (n + 1) / 2) pos = n;
  }
  for (long long i = 1; i <= n; ++i) {
    if (a[i] == 0) {
      long long res, res1;
      cout << "? " << pos << " " << i << endl;
      cin >> res;
      cout << "? " << i << " " << pos << endl;
      cin >> res1;
      if (res > res1)
        a[i] = (res1 == 0 ? a[pos] : res1) + a[pos];
      else
        a[i] = res1;
    }
  }
  cout << "! ";
  for (long long i = 1; i <= n; ++i) cout << a[i] << " ";
}