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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
const long long int mod = 1e9 + 7;
const long long int pinf = ((long long int)2e18);
const long long int ninf = ((long long int)-2e18);
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
long long int _pow(long long int a, long long int b) {
  long long int ans = 1;
  while (b) {
    if (b % 2) ans = (ans * a);
    b /= 2;
    a = (a * a);
  }
  return ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  {
    long long int s, x, y, n, i, j, p = 1, flag = 1, ans = INT_MAX;
    cin >> n;
    vector<long long int> a(n, -1), b(n);
    long long int p1 = 1, p2 = 2;
    if (n == 1) {
      cout << "! 1";
      exit(0);
    }
    for (long long int i = 2; i < n + 1; i++) {
      long long int k1, k2;
      cout << "? " << p1 << " " << p2 << '\n';
      cout.flush();
      cin >> k1;
      cout.flush();
      cout << "? " << p2 << " " << p1 << '\n';
      cout.flush();
      cin >> k2;
      if (k1 > k2) {
        a[p1 - 1] = k1;
        p1 = i + 1;
      } else {
        a[p2 - 1] = k2;
        p2 = i + 1;
      }
      cout.flush();
    }
    cout << "! ";
    for (long long int i = 0; i < n; i++) {
      if (a[i] == -1) a[i] = n;
      cout << a[i] << " ";
    }
  }
  return 0;
}