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;
typedef long long int i64;
const i64 INF = 1e18;
i64 mod = 1e9 + 7;
i64 T, n, m, k;
vector<vector<i64>> dp;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(16);
  T = 1;
  for (i64 tt = 0; tt < T; ++tt) {
    cin >> n;
    queue<i64> q;
    for (i64 i = 1; i <= n; ++i) {
      q.push(i);
    }
    vector<i64> a(n + 1, 0);
    set<i64> sx;
    for (i64 i = 1; i <= n; ++i) {
      sx.emplace(i);
    }
    while (q.size() > 1) {
      i64 x = q.front();
      q.pop();
      i64 y = q.front();
      q.pop();
      cout << "? " << x << ' ' << y << endl;
      cout.flush();
      i64 x1;
      cin >> x1;
      cout << "? " << y << ' ' << x << endl;
      cout.flush();
      i64 y1;
      cin >> y1;
      if (x1 > y1) {
        a[x] = x1;
        q.push(y);
        sx.erase(x1);
      } else {
        a[y] = y1;
        q.push(x);
        sx.erase(y1);
      }
    }
    i64 z = q.front();
    a[z] = *sx.begin();
    cout << "! ";
    for (i64 i = 1; i <= n; ++i) {
      cout << a[i] << ' ';
    }
  }
  return 0;
}
