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>
const int kN = 1e4 + 5;
int n, vis[kN], a[kN], ans[kN];
int query(int x, int y) {
  printf("? %d %d\n", a[x], a[y]);
  fflush(stdout);
  int res;
  scanf("%d", &res);
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) a[i] = i;
  int cur = n;
  while (true) {
    if (cur == 1) {
      for (int i = 1; i <= n; ++i)
        if (!vis[i]) {
          vis[i] = a[1];
          break;
        };
      break;
    } else if (cur == 2) {
      int t1 = query(1, 2), t2 = query(2, 1);
      if (t1 > t2) std::swap(a[1], a[2]);
      vis[std::max(t1, t2)] = a[2];
      for (int i = 1; i <= n; ++i)
        if (!vis[i]) vis[i] = a[1];
      break;
    } else {
      for (int i = 1, j = cur; i < j; ++i, --j) {
        int t1 = query(i, j), t2 = query(j, i);
        if (t1 > t2) std::swap(a[i], a[j]);
        vis[std::max(t1, t2)] = a[j];
      }
      cur = (cur - 1) / 2 + 1;
    }
  }
  printf("! ");
  for (int i = 1; i <= n; ++i) ans[vis[i]] = i;
  for (int i = 1; i <= n; ++i) printf("%d ", ans[i]);
  return 0;
}
