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;
template <class T>
void chmax(T &a, T b) {
  if (a < b) a = b;
}
template <class T>
void chmin(T &a, T b) {
  if (a > b) a = b;
}
constexpr int INF = 1000000000;
constexpr long long llINF = 1000000000000000000;
constexpr int mod = 1000000007;
constexpr double eps = 1e-8;
const double pi = acos(-1);
int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};
int Random(int mi, int ma) {
  random_device rnd;
  mt19937 mt(rnd());
  uniform_int_distribution<int> engine(mi, ma);
  return engine(mt);
}
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int lcm(int a, int b) { return a / gcd(a, b) * b; }
bool prime(int a) {
  if (a == 1) return false;
  for (int i = 2; i * i <= a; i++) {
    if (a % i == 0) return false;
  }
  return true;
}
int modpow(int a, int b) {
  if (b == 0) return 1;
  if (b % 2) return modpow(a, b - 1) * a % mod;
  int memo = modpow(a, b / 2);
  return memo * memo % mod;
}
vector<int> kaijo, invkaijo;
void init_fact(int n) {
  kaijo.resize(n + 1);
  invkaijo.resize(n + 1);
  kaijo[0] = 1;
  for (int i = 1; i <= n; i++) {
    kaijo[i] = kaijo[i - 1] * i;
    kaijo[i] %= mod;
  }
  for (int i = 0; i < n + 1; i++) invkaijo[i] = modpow(kaijo[i], mod - 2);
}
int comb(int a, int b) {
  if (a < b) return 0;
  if (a < 0 || b < 0) return 0;
  return kaijo[a] * invkaijo[a - b] % mod * invkaijo[b] % mod;
}
int inv(int x) {
  x = modpow(x, mod - 2);
  return x;
}
void solve() {
  int n;
  cin >> n;
  vector<int> ans(n);
  vector<bool> used(n + 1, false);
  int r = 0;
  for (int i = 1; i < n; i++) {
    int a, b;
    cout << "? " << r + 1 << " " << i + 1 << endl;
    cin >> a;
    cout << "? " << i + 1 << " " << r + 1 << endl;
    cin >> b;
    if (a > b) {
      ans[r] = a;
      used[a] = true;
      r = i;
    } else {
      ans[i] = b;
      used[b] = true;
      r = r;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (!used[i]) {
      ans[r] = i;
    }
  }
  cout << "! ";
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
  cout << endl;
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  solve();
  return 0;
}