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;
const long long INF = 1e18 + 5;
const long long MAX_N = 1e5 + 5;
const long long MODD = 1e9 + 7;
const long double PI = 3.14159265358979323846;
long long max(long long a, long long b) { return a > b ? a : b; }
long long min(long long a, long long b) { return a < b ? a : b; }
double max(double a, double b) { return a > b ? a : b; }
double min(double a, double b) { return a < b ? a : b; }
long long modd(long long a) {
  if (a < 0LL) return a * -1LL;
  return a;
}
long double modd(long double a) {
  if (a < 0.0) return a * -1.0;
  return a;
}
long long modulo(long long a) { return (a + MODD) % MODD; }
long long power(long long k, long long p) {
  if (p == 0) return 1LL;
  long long P = power(k, p / 2LL);
  P = (P * P) % MODD;
  if (p % 2 == 1) P = (P * k) % MODD;
  return P;
}
struct p_str {
  long long a, b;
};
void solve() {
  long long n;
  cin >> n;
  queue<long long> Q;
  for (long long i = 1; i <= n; i++) {
    Q.push(i);
  }
  bool seen[n + 5];
  memset(seen, 0, sizeof(seen));
  long long qij, qji;
  long long ans[n + 5];
  while (Q.size() > 1) {
    long long i = Q.front();
    Q.pop();
    long long j = Q.front();
    Q.pop();
    cout << "? " << i << " " << j << endl;
    fflush(stdout);
    cin >> qij;
    cout << "? " << j << " " << i << endl;
    fflush(stdout);
    cin >> qji;
    if (qij < qji) {
      ans[j] = qji;
      seen[qji] = 1;
      Q.push(i);
    } else {
      ans[i] = qij;
      seen[qij] = 1;
      Q.push(j);
    }
  }
  for (long long i = 1; i <= n; i++) {
    if (seen[i] == 0) {
      ans[Q.front()] = i;
      break;
    }
  }
  cout << "! ";
  for (long long i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
  fflush(stdout);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  for (long long i = 0; i < t; i++) {
    solve();
  }
  return 0;
}