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>
#pragma GCC optimize("Ofast")
using namespace std;
long long max(long long a, long long b) {
  if (a > b) {
    return a;
  } else {
    return b;
  }
}
long long min(long long a, long long b) {
  if (a < b) {
    return a;
  } else {
    return b;
  }
}
long long power(long long b, long long e) {
  if (e == 0) return 1;
  if (e % 2)
    return ((b * power((b) * (b), (e - 1) / 2)));
  else
    return power((b) * (b), e / 2);
}
long long modpower(long long b, long long e, long long q) {
  long long MOD = q;
  if (e == 0) return 1;
  if (e % 2)
    return ((b % MOD) * modpower((b % MOD) * (b % MOD), (e - 1) / 2, q)) % MOD;
  else
    return modpower((b % MOD) * (b % MOD), e / 2, q) % MOD;
}
void dpv(vector<long long> v) {
  for (long long i = 0; i < v.size(); i++) {
    cout << v[i] << " ";
  }
  cout << endl;
}
void dpv(vector<pair<long long, long long> > v) {
  for (long long i = 0; i < v.size(); i++) {
    cout << v[i].first << " " << v[i].second << endl;
  }
}
void dpv(set<long long> v) {
  for (auto i : v) {
    cout << i << " ";
  }
  cout << endl;
}
void oblivious() {
  long long n;
  cin >> n;
  if (n == 1) {
    cout << "! 1" << endl;
    fflush(stdout);
    return;
  } else {
    long long a[n];
    long long l = 0, r = n - 1;
    long long ans = 0;
    long long cnt = n * (n + 1) / 2;
    while (l < r) {
      long long x1, x2;
      cout << "? " << l + 1 << " " << r + 1 << endl;
      fflush(stdout);
      cin >> x1;
      cout << "? " << r + 1 << " " << l + 1 << endl;
      fflush(stdout);
      cin >> x2;
      if (x2 > x1) {
        a[r] = x2;
        ans += x2;
        r--;
      } else {
        a[l] = x1;
        ans += x1;
        l++;
      }
    }
    a[l] = cnt - ans;
    cout << "!";
    for (long long i = 0; i < n; i++) {
      cout << " " << a[i];
    }
    cout << endl;
    fflush(stdout);
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    oblivious();
  }
  return 0;
}
