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;
using ull = unsigned long long;
using ll = long long;
int t, n;
int _p[10005];
int p[10005];
int b[10005];
int cnt[10005];
vector<int> vec[10005];
map<int, int> ans[10005];
int tot = 0;
int ask(int x, int y) {
  if (ans[x].count(y)) {
    return ans[x][y];
  }
  ++tot;
  assert(tot <= 2 * n);
  cout << "? " << x << ' ' << y << endl;
  cout.flush();
  int v;
  cin >> v;
  return ans[x][y] = v;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> n;
  if (n == 1) {
    cout << "! 1" << endl;
    cout.flush();
    return 0;
  }
  int pos = 0;
  for (int i = 1; i <= n; i += 2) {
    if ((n & 1) && i == n) {
      --i;
    }
    int x = ask(i, i + 1), y = ask(i + 1, i);
    if (x > y) {
      p[i] = x;
      if (x >= p[pos]) {
        pos = i;
      }
    } else {
      p[i + 1] = y;
      if (y >= p[pos]) {
        pos = i + 1;
      }
    }
    if ((n & 1) && i == n - 1) {
      break;
    }
  }
  assert(pos != -1);
  map<int, bool> was;
  for (int i = 1; i <= n; ++i) {
    if (p[i] != 0) {
      was[p[i]] = true;
    }
    if (i == pos || p[i] != 0) {
      continue;
    }
    int x = ask(i, pos);
    vec[x].push_back(i);
  }
  for (int i = 0; i < p[pos]; ++i) {
    if ((int)vec[i].size() == 1) {
      int x = (i == 0 ? p[pos] : i) + p[pos];
      if (x > n || was[x]) {
        p[vec[i][0]] = i;
      } else {
        p[vec[i][0]] = x;
      }
      continue;
    }
    if (!vec[i].empty() && (int)vec[i].size() != 2) {
      abort();
    }
    if (!vec[i].empty()) {
      int x = ask(vec[i][0], vec[i][1]), y = ask(vec[i][1], vec[i][0]);
      if (x < y) {
        p[vec[i][0]] = p[pos] + max(x, y);
        p[vec[i][1]] = max(x, y);
      } else {
        p[vec[i][1]] = p[pos] + max(x, y);
        p[vec[i][0]] = max(x, y);
      }
    }
  }
  cout << '!';
  for (int i = 1; i <= n; ++i) {
    cout << ' ' << p[i];
  }
  cout << endl;
  cout.flush();
  return 0;
}