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;
int wait;
const double pi =
    3.14159265358979323846264338327950288419716939937510582097494459;
const double EPS = 1e-12;
const int meminf = 0x3f;
const int MAXN = 1e5 + 10;
const long long MOD = 1e9 + 7;
const long long Mod = 1e9 + 7;
const long long mod = 1e9 + 7;
const long long modd = 998244353;
const int MAX = 1e7 + 5;
const int oo = 2e9 + 10;
const long long OO = 1e18 + 10;
const int N = 1e5 + 5;
const int logN = (int)(log2(N));
int dx[] = {0, 1, 0, -1, -1, -1, 1, 1};
int dy[] = {1, 0, -1, 0, -1, 1, -1, 1};
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
long long lcm(long long a, long long b) { return ((a * b) / gcd(a, b)); }
int main() {
  map<pair<int, int>, int> mods;
  int n;
  cin >> n;
  int mx = 0;
  int tmp;
  vector<int> vec(n, -1);
  for (int i = 1; i < n; i++) {
    cout << "? " << i + 1 << " " << mx + 1 << endl;
    cout.flush();
    cin >> tmp;
    mods[{i, mx}] = tmp;
    cout << "? " << mx + 1 << " " << i + 1 << endl;
    cout.flush();
    cin >> tmp;
    mods[{mx, i}] = tmp;
    if (mods[{mx, i}] > mods[{i, mx}]) {
      vec[mx] = mods[{mx, i}];
      mx = i;
    } else {
      vec[i] = mods[{i, mx}];
    }
  }
  vec[mx] = n;
  cout << "! ";
  for (auto x : vec) cout << x << " ";
  cout << "\n";
}