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 checkpoint1() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
bool comp(pair<long long, long long> &a, pair<long long, long long> &b) {
  if (a.first != b.first)
    return a.first < b.first;
  else {
    if (a.second > b.second)
      return b.second < a.second;
    else if (a.second < b.second)
      return a.second < b.second;
  }
  return a.first > b.first;
}
long long powe(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) {
      res *= a;
      res %= 1000000007;
    }
    a *= a;
    a %= 1000000007;
    b >>= 1;
  }
  return res;
}
int main() {
  checkpoint1();
  long long n;
  cin >> n;
  long long ans[n + 1];
  bool vis[n + 1];
  memset(vis, 0, sizeof(vis));
  memset(ans, 0, sizeof(ans));
  set<long long> s;
  for (long long i = 1; i <= n; i++) s.insert(i);
  while (s.empty() == 0) {
    if (s.size() == 1) break;
    long long i1, i2;
    long long i = *s.begin();
    s.erase(s.begin());
    long long j = *s.begin();
    s.erase(s.begin());
    cout << "? " << i << " " << j << endl;
    cin >> i1;
    cout.flush();
    cout << "? " << j << " " << i << endl;
    cin >> i2;
    cout.flush();
    if (i1 > i2) {
      ans[i] = i1;
      vis[i1] = 1;
      s.insert(j);
    } else {
      ans[j] = i2;
      vis[i2] = 1;
      s.insert(i);
    }
  }
  long long fin = *s.begin();
  long long any = -1;
  for (long long ii = 1; ii <= n; ii++) {
    if (vis[ii] == 0) {
      any = ii;
      break;
    }
  }
  ans[fin] = any;
  cout << "! ";
  for (long long i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
  return 0;
}
