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;
using ull = unsigned long long;
using ld = long double;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
const ll N = 1e7 + 10;
const ll INF = 1e18;
const ll M = 1e3 + 1;
const ll L = 31;
const ll mod = 1e9 + 7;
ll ask(ll i, ll j) {
  cout << "? ";
  cout << i << " " << j;
  cout << endl;
  cout.flush();
  ll x;
  cin >> x;
  return x;
}
ll solve() {
  ll n;
  cin >> n;
  ll arr[n];
  for (ll i = 1; i <= n; i++) {
    arr[i] = -1;
  }
  ll brr[n];
  for (ll i = 1; i <= n; i++) {
    brr[i] = -1;
  }
  ll i = 1, j = 2;
  while (i <= n && j <= n) {
    ll a = ask(i, j);
    ll b = ask(j, i);
    if (a > b) {
      arr[i] = a;
      brr[a] = 1;
      i = j;
    } else {
      arr[j] = b;
      brr[b] = 1;
    }
    j++;
  }
  ll h;
  for (ll i = 1; i <= n; i++) {
    if (brr[i] == -1) {
      h = i;
      break;
    }
  }
  for (ll i = 1; i <= n; i++) {
    if (arr[i] == -1) {
      arr[i] = h;
    }
  }
  cout << "! ";
  for (ll i = 1; i <= n; i++) {
    cout << arr[i] << " ";
  }
  return 0;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  solve();
  return 0;
}