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;
long long MOD = 1000000007;
long long bigmod(long long a, long long b) {
  if (b == 0) return 1;
  long long x = bigmod(a, b / 2);
  x = (x * x) % MOD;
  if (b % 2 == 1) x *= a;
  return x % MOD;
}
int t;
int n;
int ans[10005];
bool vis[10005];
vector<int> kh(vector<int> v) {
  for (int i = 0; i <= v.size(); i++) vis[i] = 0;
  for (int i = 0; i < v.size(); i += 2) {
    if (i + 1 >= v.size()) break;
    int x = v[i];
    int y = v[i + 1];
    cout << "? " << x << " " << y << endl;
    int rem1;
    cin >> rem1;
    cout << "? " << y << " " << x << endl;
    int rem2;
    cin >> rem2;
    if (rem1 > rem2) {
      ans[x] = rem1;
      vis[i] = 1;
    } else {
      ans[y] = rem2;
      vis[i + 1] = 1;
    }
  }
  vector<int> ret;
  for (int i = 0; i < v.size(); i++) {
    if (!vis[i]) ret.push_back(v[i]);
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  vector<int> v;
  for (int i = 1; i <= n; i++) v.push_back(i);
  while (v.size() > 1) {
    v = kh(v);
  }
  ans[v[0]] = n;
  cout << "! ";
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
  return 0;
}