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>
using namespace std;
int power(long long int x, long long int y, long long int m) {
  if (y == 0) return 1;
  long long int p = power(x, y / 2, m) % m;
  p = (p * p) % m;
  return (y % 2 == 0) ? p : (x * p) % m;
}
int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
vector<long long int> getDivisors(long long int n) {
  vector<long long int> vec;
  for (int i = 1; i <= sqrtl(n) + 1; i++) {
    if (n % i == 0) {
      if (n / i == i) {
        vec.push_back(i);
      } else {
        vec.push_back(i);
        vec.push_back(n / i);
      }
    }
  }
  return vec;
}
pair<long long int, long long int> fib(long long int n) {
  if (n == 0) return {0, 1};
  auto p = fib(n >> 1);
  long long int c = p.first * (2 * p.second - p.first);
  long long int d = p.first * p.first + p.second * p.second;
  if (n & 1)
    return {d, c + d};
  else
    return {c, d};
}
long long int fact(long long int a, long long int b, long long int c) {
  long long int f = 1;
  long long int flag = 1;
  for (long long int i = a; i > b; i--) {
    f *= i;
    if (f % c == 0 && flag == 1) {
      f /= c;
      flag = 0;
    }
  }
  return f;
}
int main() {
  long long int t = 1;
  while (t--) {
    int n;
    cin >> n;
    vector<int> v(n, 0);
    queue<int> q;
    for (int i = 0; i < n; i++) {
      q.push(i);
    }
    int a, b;
    int ans1 = 0, ans2 = 0;
    while (q.size() > 1) {
      a = q.front();
      q.pop();
      b = q.front();
      q.pop();
      cout << "? " << a + 1 << " " << b + 1 << endl;
      cout.flush();
      cin >> ans1;
      cout << "? " << b + 1 << " " << a + 1 << endl;
      cout.flush();
      cin >> ans2;
      if (ans1 > ans2) {
        v[a] = ans1;
        q.push(b);
      } else {
        v[b] = ans2;
        q.push(a);
      }
    }
    a = q.front();
    v[a] = n;
    cout << "! ";
    for (int i = 0; i < n; i++) {
      cout << v[i] << " ";
    }
    cout << endl;
    cout.flush();
  }
  return 0;
}
