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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long hcf(long long n1, long long n2) {
  if (n2 != 0)
    return hcf(n2, n1 % n2);
  else
    return n1;
}
long long fact(long long n) {
  long long i = 1, co = 1;
  for (i = 1; i <= n; i++) co *= i;
  return co;
}
long long isprime(long long n) {
  if (n <= 1) return 0;
  if (n <= 3) return 1;
  if (n % 2 == 0 || n % 3 == 0) return 0;
  for (long long i = 5; i <= sqrt(n); i += 6) {
    if (n % i == 0 || n % (i + 2) == 0) return 0;
  }
  return 1;
}
long long sod(long long n) {
  long long sum = 0;
  while (n) {
    sum += n % 10;
    n /= 10;
  }
  return sum;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  t = 1;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> ans(n, 0);
    if (n == 1) {
      ans[0] = 1;
    } else {
      long long x = 0, y = 1;
      for (int i = 0; i <= n - 2; i++) {
        long long res1, res2;
        cout << '?' << ' ' << x + 1 << ' ' << y + 1 << endl;
        cin >> res1;
        cout << '?' << ' ' << y + 1 << ' ' << x + 1 << endl;
        cin >> res2;
        if (res1 < res2) {
          swap(res1, res2);
          swap(x, y);
        }
        ans[x] = res1;
        for (int i = 0; i <= n - 1; i++) {
          if (ans[i] == 0 && y != i) x = i;
        }
      }
      ans[y] = n;
    }
    cout << '!' << ' ';
    for (auto it : ans) cout << it << ' ';
    cout << endl;
  }
  return 0;
}
