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 ll = long long;
using pii = pair<int, int>;
using db = double;
using vi = vector<int>;
const int inf = 0x3f3f3f3f;
const db eps = 1e-8;
const int mod = 1e9 + 7;
ll qpow(ll a, ll b) {
  ll ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ret;
}
const int maxn = 1e4 + 100;
int vis[maxn];
set<int> st;
int main() {
  int n;
  cin >> n;
  int cnt = n;
  for (int i = 1; i <= n; i++) st.insert(i);
  while (cnt != 1) {
    int tmp1, tmp2;
    if (st.size() == 1) break;
    auto i = st.begin();
    auto j = i;
    j++;
    cout << "? " << *i << " " << *j << endl;
    cout.flush();
    cin >> tmp1;
    cout << "? " << *j << " " << *i << endl;
    cout.flush();
    cin >> tmp2;
    if (tmp1 > tmp2) {
      vis[*i] = tmp1;
      st.erase(*i);
    } else {
      vis[*j] = tmp2;
      st.erase(*j);
    }
    cnt--;
  }
  cout << "! ";
  cout.flush();
  for (int i = 1; i <= n; i++) {
    if (vis[i] == 0) {
      cout << n;
      cout.flush();
    } else {
      cout << vis[i];
      cout.flush();
    }
    if (i != n) {
      cout << " ";
      cout.flush();
    }
  }
  cout.flush();
}