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;
map<long long, long long> vis;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long T = 1;
  while (T--) {
    long long n;
    cin >> n;
    long long st = 1, en = n;
    long long a, b;
    long long ar[n + 2];
    long long sz = 0;
    while (st < en) {
      cout << "? " << st << " " << en << endl;
      cin >> a;
      cout << "? " << en << " " << st << endl;
      cin >> b;
      if (a < b) {
        ar[en] = b;
        en = en - 1;
        sz++;
      } else if (b < a) {
        ar[st] = a;
        st = st + 1;
        sz++;
      }
      if (sz == n) break;
    }
    long long num = -1;
    for (long long i = 1; i <= n; i++) vis[ar[i]] = 1;
    for (long long i = 1; i <= n; i++) {
      if (vis[i] == 0) {
        num = i;
        break;
      }
    }
    if (num != -1) ar[st] = num;
    cout << "! ";
    for (long long i = 1; i <= n; i++) cout << ar[i] << " ";
  }
}