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;
long long int power(long long int a, long long int b,
                    long long int m = 1000000007) {
  a %= m;
  long long int r = 1;
  while (b) {
    if (b & 1) r = r * a % m;
    b >>= 1;
    a = a * a % m;
  }
  return r;
}
long long int inv(long long int a, long long int m = 1000000007) {
  return power(a, m - 2, m);
}
long long int mult(long long int a, long long int b,
                   long long int m = 1000000007) {
  return ((a % m) * (b % m) % m);
}
bool isTwo(long long int n) { return n && (!(n & (n - 1))); }
long long int dist = 0, dia = 0, x = -1, mx = 0;
vector<bool> vis(500005);
void solve(long long int u, vector<vector<long long int>> &adj, long long int p,
           long long int cnt) {
  if (cnt > mx) {
    x = u;
    mx = cnt;
  }
  for (long long int i = 0; i < adj[u].size(); i++) {
    if (adj[u][i] != p) solve(adj[u][i], adj, u, cnt + 1);
  }
}
void solve2(long long int u, vector<vector<long long int>> &adj,
            long long int p, long long int cnt) {
  if (cnt > mx) {
    mx = cnt;
    dia = cnt;
  }
  for (long long int i = 0; i < adj[u].size(); i++) {
    if (adj[u][i] != p) solve2(adj[u][i], adj, u, cnt + 1);
  }
}
void dfs(long long int u, vector<vector<long long int>> &adj, long long int d,
         long long int b, long long int p) {
  if (u == b) {
    dist = d;
    return;
  }
  for (long long int i = 0; i < adj[u].size(); i++) {
    if (adj[u][i] != p) dfs(adj[u][i], adj, d + 1, b, u);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long int n, i;
  cin >> n;
  vector<long long int> ans(n);
  set<long long int> s;
  for (long long int i = 1; i <= n; i++) {
    s.insert(i);
  }
  for (i = 1; i <= n && (long long int)s.size() >= 2; i++) {
    long long int f1 = *s.begin();
    s.erase(s.begin());
    long long int f2 = *s.begin();
    s.erase(s.begin());
    long long int x, y;
    cout << "? " << f1 << ' ' << f2 << endl;
    cout.flush();
    cin >> x;
    cout << "? " << f2 << ' ' << f1 << endl;
    cout.flush();
    cin >> y;
    if (x > y) {
      ans[f1 - 1] = x;
      s.insert(f2);
    } else {
      ans[f2 - 1] = y;
      s.insert(f1);
    }
  }
  vector<bool> visited(n);
  for (long long int i = 0; i < n; i++) {
    if (ans[i] > 0) visited[ans[i] - 1] = true;
  }
  long long int r = 0;
  for (long long int i = 0; i < n; i++) {
    if (visited[i] == false) r = i;
  }
  for (long long int i = 0; i < n; i++) {
    if (ans[i] == 0) ans[i] = r + 1;
  }
  cout << "! ";
  for (i = 0; i < ans.size(); i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
  cout.flush();
  long long int verdict;
  cin >> verdict;
}
