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>
using namespace std;
using ll = long long;
const int N = 1e6 + 10;
const int inf = 1e9;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int main() {
  int n;
  cin >> n;
  auto ask = [&](int i, int j) {
    cout << "? " << i + 1 << " " << j + 1 << endl;
    int x;
    cin >> x;
    return x;
  };
  vector<int> ans(n);
  set<int> hz;
  for (int i = 0; i < n; i++) hz.insert(i);
  while (hz.size() > 1) {
    auto i1 = *hz.begin();
    hz.erase(i1);
    auto i2 = *hz.begin();
    hz.erase(i2);
    int x1 = ask(i1, i2);
    int x2 = ask(i2, i1);
    if (x1 > x2) {
      ans[i1] = x1;
      hz.insert(i2);
    } else {
      ans[i2] = x2;
      hz.insert(i1);
    }
  }
  ans[*hz.begin()] = n * (n + 1) / 2 - accumulate(ans.begin(), ans.end(), 0);
  cout << "!"
       << " ";
  for (auto x : ans) cout << x << " ";
  cout << endl;
}