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;
template <class A, class B>
ostream &operator<<(ostream &out, const pair<A, B> &a) {
  return out << "(" << a.first << "," << a.second << ")";
}
template <class A>
ostream &operator<<(ostream &out, const vector<A> &a) {
  for (const A &it : a) out << it << " ";
  return out;
}
template <class A, class B>
istream &operator>>(istream &in, pair<A, B> &a) {
  return in >> a.first >> a.second;
}
template <class A>
istream &operator>>(istream &in, vector<A> &a) {
  for (A &i : a) in >> i;
  return in;
}
ifstream cinn("in.txt");
ofstream coutt("out.txt");
long long poww(const long long &a, long long b,
               const long long &m = 1000000007) {
  if (b == 0) return 1;
  long long first = poww(a, b / 2, m);
  first = first * first % m;
  if (b & 1) first = first * a % m;
  return first;
}
long long gcd(long long a, long long b) {
  if (a > b) swap(a, b);
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long rint(long long l, long long r) { return rand() % (r - l + 1) + l; }
signed main() {
  long long i, j, first, n;
  cin >> n;
  set<long long> st;
  set<long long> indice;
  vector<long long> ans(n + 1);
  for (i = 1; i < n + 1; i++) {
    st.insert(i);
    indice.insert(i);
  }
  while (indice.size() > 1) {
    vector<long long> v;
    for (auto aa : indice) {
      v.push_back(aa);
    }
    long long sz = v.size();
    for (i = 0; i + 1 < sz; i += 2) {
      long long a, b;
      cout << "? " << v[i] << " " << v[i + 1] << '\n';
      cout.flush();
      cin >> a;
      cout << "? " << v[i + 1] << " " << v[i] << '\n';
      cout.flush();
      cin >> b;
      if (a > b) {
        ans[v[i]] = a;
        indice.erase(v[i]);
        st.erase(a);
      } else {
        ans[v[i + 1]] = b;
        indice.erase(v[i + 1]);
        st.erase(b);
      }
    }
  }
  if (indice.size() == 1) {
    first = *indice.begin();
    i = *st.begin();
    ans[first] = i;
  }
  cout << "! ";
  for (i = 1; i < n + 1; i++) {
    cout << ans[i] << " ";
  }
  cout << '\n';
  cout.flush();
  return 0;
}
