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;
void solve() {
  long long int a;
  while (cin >> a && a != -1) {
    if (a == -1) exit(0);
    if (a == 1) {
      cout << "! 1" << endl;
      continue;
    }
    long long int in = 2;
    vector<int> v(a + 1);
    for (long long int i = 1; i <= a && in <= a;) {
      while (v[i] != 0) i++;
      if (i > a) break;
      if (i == in) in++;
      if (in > a) break;
      long long int x, y;
      cout << "? " << i << ' ' << in << endl;
      cout.flush();
      cin >> x;
      cout << "? " << in << ' ' << i << endl;
      cout.flush();
      cin >> y;
      if (x > y) {
        v[i] = x;
        i++;
      } else {
        v[in] = y;
        in++;
      }
    }
    vector<int> vv;
    for (long long int i = 1; i <= a; i++) {
      if (v[i] == 0) vv.push_back(i);
    }
    if (vv.size() > 1) {
      long long int i = vv[0], in = vv[1];
      long long int x, y;
      cout << "? " << i << ' ' << in << endl;
      cout.flush();
      cin >> x;
      cout << "? " << in << ' ' << i << endl;
      cout.flush();
      cin >> y;
      if (x > y) {
        v[i] = x;
        i++;
      } else {
        v[in] = y;
        in++;
      }
    }
    cout << "! ";
    for (long long int i = 1; i <= a; i++) {
      if (v[i] == 0) v[i] = a;
      cout << v[i] << ' ';
    }
    cout << endl;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  solve();
  return 0;
}
