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;
const long double pi = 3.141592653589793238462643383279;
const long long MOD = 1e9 + 7;
const long long INF = 1e9;
const long double eps = 0.000000000000001;
void boot_up() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.setf(ios::fixed);
  cout.precision(15);
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
signed main() {
  boot_up();
  long long n, d;
  cin >> n;
  if (n == 1) {
    cout << "! " << 1 << endl;
    return 0;
  }
  vector<long long> b(n + 1, 0), nums(n + 1, 0);
  long long f, s;
  cout << "? " << 1 << " " << 2 << endl;
  cin >> f;
  cout << "? " << 2 << " " << 1 << endl;
  cin >> s;
  if (f > s) {
    d = 2;
    b[1] = f;
  } else {
    d = 1;
    b[2] = s;
  }
  for (long long i = 3; i <= n; i++) {
    cout << "? " << i << " " << d << endl;
    cin >> f;
    cout << "? " << d << " " << i << endl;
    cin >> s;
    if (f > s) {
      b[i] = f;
    } else {
      b[d] = s;
      d = i;
    }
  }
  for (long long i = 1; i <= n; i++) {
    nums[b[i]] = 1;
  }
  long long numb;
  for (long long i = 1; i <= n; i++) {
    if (nums[i] == 0) {
      numb = i;
    }
  }
  b[d] = numb;
  cout << "! ";
  for (long long i = 1; i <= n; i++) {
    cout << b[i] << " ";
  }
  return 0;
}
