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;
template <class T>
ostream &operator<<(ostream &os, vector<T> V) {
  os << "[ ";
  for (auto v : V) os << v << " ";
  os << "]";
  return os;
}
template <class T>
ostream &operator<<(ostream &os, set<T> S) {
  os << "{ ";
  for (auto s : S) os << s << " ";
  return os << "}";
}
template <class L, class R>
ostream &operator<<(ostream &os, map<L, R> M) {
  os << "{ ";
  for (auto m : M) os << "(" << m.first << ":" << m.second << ") ";
  return os << "}";
}
template <class L, class R>
ostream &operator<<(ostream &os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
const long long N = 10005;
long long a[N];
bool vis[N];
long long ask(long long x, long long y) {
  cout << "? " << x << ' ' << y << endl;
  long long c;
  cin >> c;
  return c;
}
int main() {
  long long n;
  cin >> n;
  memset(a, -1, sizeof(a));
  while (1) {
    vector<long long> v;
    for (long long i = 1; i < n + 1; i++) {
      if (a[i] == -1) v.push_back(i);
    }
    if (v.empty()) break;
    if (v.size() == 1) {
      for (long long i = 1; i < n + 1; i++)
        if (a[i] != -1) vis[a[i]] = true;
      for (long long i = 1; i < n + 1; i++) {
        if (!vis[i]) {
          a[v[0]] = i;
          break;
        }
      }
      break;
    }
    for (long long i = 0; i < v.size(); i++) {
      if (i == (int)v.size() - 1) break;
      long long k1 = ask(v[i], v[i + 1]);
      long long k2 = ask(v[i + 1], v[i]);
      if (k1 > k2)
        a[v[i]] = k1;
      else
        a[v[i + 1]] = k2;
      i++;
    }
  }
  cout << "! ";
  for (long long i = 1; i < n + 1; i++) cout << a[i] << ' ';
  cout << endl;
  return 0;
}
