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 int N = 1e6 + 9;
const long long int inf = 1e14;
const long long int MOD = 1e9 + 7;
const long long int mod = 1e9 + 7;
const double PI = acos(-1);
long long int powm(long long int a, long long int b) {
  a = a % mod;
  long long int res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
int main() {
  long long int t = 1;
  while (t--) {
    long long int n;
    cin >> n;
    set<long long int> a, b;
    for (int i = 1; i <= n; i++) {
      a.insert(i);
      b.insert(i);
    }
    long long int an[n];
    while (a.size() != 1) {
      long long int v1 = *a.begin(), v2 = *(++a.begin());
      long long int x1, x2;
      cout << "? " << v1 << " " << v2 << '\n';
      cin >> x1;
      cout << "? " << v2 << " " << v1 << '\n';
      cin >> x2;
      if (x2 > x1) {
        an[v2 - 1] = x2;
        a.erase(v2);
        b.erase(x2);
      } else {
        an[v1 - 1] = x1;
        a.erase(v1);
        b.erase(x1);
      }
    }
    long long int ind = *a.begin(), va = *b.begin();
    an[ind - 1] = va;
    cout << "! ";
    for (int i = 0; i < n; i++) {
      cout << an[i] << " ";
    }
    cout << '\n';
  }
}