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;
long long int mod1 = pow(10, 9) + 7;
int mod2 = 998244353;
long long int inf = 1e9;
long long int maxxy = 1e6 + 9;
const double PI = 3.141592653589793238;
unsigned long long int sumcons(unsigned long long int a,
                               unsigned long long int b) {
  if (a == 0) {
    return (b * (b - 1)) / 2;
  } else {
    unsigned long long int zzz = (b * (b + 1)) / 2;
    zzz += ((a - 1) * b);
    return zzz;
  }
}
long long int power(long long int x, long long int y, long long int pp) {
  long long int res = 1;
  x = x % pp;
  while (y > 0) {
    if (y & 1) res = (res * x) % pp;
    y = y >> 1;
    x = (x * x) % pp;
  }
  return res;
}
long long int modInverse(long long int n, long long int pp) {
  return power(n, pp - 2, pp);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  vector<long long int> ans(n + 1, 0);
  if (n == 1) {
    cout << "! 1" << endl;
  } else {
    long long int moves = 0;
    long long int l = 1;
    long long int r = n;
    long long int q;
    long long int qq;
    long long int cc = n / 2;
    if (n % 2 == 1) {
      cc++;
    }
    vector<pair<pair<long long int, long long int>, long long int>> v;
    while (l < r) {
      long long int a = 0, b = 0;
      cout << "? " << l << " " << r << endl;
      cin >> q;
      a = q;
      v.push_back({{l, r}, a});
      moves++;
      if (a == 0) {
        r--;
      } else if (a >= cc) {
        ans[l] = a;
        l++;
      } else {
        cout << "? " << r << " " << l << endl;
        cin >> qq;
        b = qq;
        v.push_back({{r, l}, b});
        if (a > b) {
          ans[l] = a;
          l++;
        } else {
          ans[r] = b;
          r--;
        }
      }
    }
    ans[l] = n;
    for (int i = 0; i < v.size(); ++i) {
      if (v[i].first.second == l) {
        ans[v[i].first.first] = v[i].second;
      }
    }
    long long int kkk;
    for (int i = 1; i <= n; ++i) {
      if (ans[i] == 0) {
        cout << "? " << i << " " << l << endl;
        cin >> kkk;
        ans[i] = kkk;
      }
    }
    cout << "! ";
    for (int i = 1; i <= n; ++i) {
      cout << ans[i] << " ";
    }
    cout << endl;
  }
  return 0;
}