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>
#pragma GCC optimize("O2")
using namespace std;
using pii = pair<int, int>;
using pll = pair<long long, long long>;
using vb = vector<bool>;
using vi = vector<int>;
using vl = vector<long long>;
using vvb = vector<vector<bool>>;
using vvi = vector<vector<int>>;
using vvl = vector<vector<long long>>;
using vpii = vector<pair<int, int>>;
using mii = map<int, int>;
template <typename... ArgTypes>
void print(ArgTypes... args);
template <typename... ArgTypes>
void input(ArgTypes &...args);
template <>
void print() {}
template <>
void input() {}
template <typename T, typename... ArgTypes>
void print(T t, ArgTypes... args) {
  cout << t;
  print(args...);
}
template <typename T, typename... ArgTypes>
void input(T &t, ArgTypes &...args) {
  cin >> t;
  input(args...);
}
int n;
vi ans;
void traverse(int now, int previous) {
  if (now > n) {
    ans[previous - 1] = n;
    return;
  }
  int a;
  int b;
  cout << "? " << now << " " << previous << endl;
  cout.flush();
  cin >> a;
  cout << "? " << previous << " " << now << endl;
  cout.flush();
  cin >> b;
  if (a > b) {
    ans[now - 1] = a;
    return traverse(now + 1, previous);
  }
  ans[previous - 1] = b;
  traverse(now + 1, now);
}
int main() {
  cin >> n;
  if (n == 1) {
    cout << "! 1" << endl;
    cout.flush();
    return 0;
  }
  ans = vi(n);
  traverse(2, 1);
  cout << "!";
  for (int i = 0; i < n; i++) cout << " " << ans[i];
  cout << endl;
  cout.flush();
}