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;
const int N = 1e4 + 10;
int a[N], vis[N];
int ask(int x, int y) {
  int k;
  printf("? %d %d\n", x, y);
  fflush(stdout), scanf("%d", &k);
  return k;
}
int main() {
  int n;
  scanf("%d", &n);
  queue<int> q;
  for (int i = 1; i <= n; i++) q.push(i);
  while (q.size() >= 2) {
    int u = q.front();
    q.pop();
    int v = q.front();
    q.pop();
    int k1 = ask(u, v), k2 = ask(v, u);
    if (k1 > k2)
      a[u] = k1, vis[k1] = 1, q.push(v);
    else
      a[v] = k2, vis[k2] = 1, q.push(u);
  }
  int t;
  for (int i = 1; i <= n; i++)
    if (!vis[i]) t = i;
  a[q.front()] = t;
  printf("! ");
  for (int i = 1; i <= n; i++) printf("%d ", a[i]);
  puts("");
  return 0;
}