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 vl = vector<long long>;
using vs = vector<string>;
using vvl = vector<vector<long long>>;
const long long INF = 1LL << 60;
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
bool compare(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first != b.first) {
    return a.first < b.first;
  } else {
    return a.second < b.second;
  }
}
bool In_map(long long y, long long x, long long h, long long w) {
  if (y < 0 || x < 0 || y >= h || x >= w) {
    return 0;
  } else {
    return 1;
  }
}
const vector<long long> dx{1, 0, -1, 0};
const vector<long long> dy{0, 1, 0, -1};
void Main() {
  long long n;
  cin >> n;
  long long now = 1;
  vector<long long> ans(n);
  for (int i = 1; i <= (n - 1); i++) {
    long long a, b;
    cout << "? " << now << " " << i + 1 << endl;
    cin >> a;
    cout << "? " << i + 1 << " " << now << endl;
    cin >> b;
    if (a > b) {
      ans[now - 1] = a;
      now = i + 1;
    } else {
      ans[i] = b;
    }
  }
  ans[now - 1] = n;
  cout << "! ";
  for (int i = 0; i < (n); i++) {
    if (i == n - 1) {
      cout << ans[i] << endl;
    } else {
      cout << ans[i] << " ";
    }
  }
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t = 1;
  while (t--) {
    Main();
  }
  return 0;
}