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 INF = 1LL << 60;
long long MOD = 1e9 + 7;
namespace output {
void pr(short x) { cout << x; }
void pr(signed x) { cout << x; }
void pr(long x) { cout << x; }
void pr(long long x) { cout << x; }
void pr(unsigned short x) { cout << x; }
void pr(unsigned x) { cout << x; }
void pr(unsigned long x) { cout << x; }
void pr(unsigned long long x) { cout << x; }
void pr(float x) { cout << x; }
void pr(double x) { cout << x; }
void pr(long double x) { cout << x; }
void pr(char x) { cout << x; }
void pr(const char* x) { cout << x; }
void pr(const string& x) { cout << x; }
void pr(bool x) { cout << x; }
template <size_t sz>
void pr(const bitset<sz>& x) {
  cout << x;
}
template <class T>
void pr(const complex<T>& x) {
  cout << x;
}
template <class T1, class T2>
void pr(const pair<T1, T2>& x);
template <class T1, class T2, class T3>
void pr(const tuple<T1, T2, T3>& x);
template <class T>
void pr(const T& x);
template <class T, class... Ts>
void pr(const T& t, const Ts&... ts) {
  pr(t);
  pr(ts...);
}
template <class T1, class T2>
void pr(const pair<T1, T2>& x) {
  pr("(", x.first, ", ", x.second, ")");
}
template <class T1, class T2, class T3>
void pr(const tuple<T1, T2, T3>& x) {
  pr("(", get<0>(x), ", ", get<1>(x), ", ", get<2>(x), ")");
}
template <class T>
void pr(const T& x) {
  pr("[");
  bool fst = 1;
  for (const auto& a : x) pr(!fst ? ", " : "", a), fst = 0;
  pr("]");
}
void print() { pr("\n"); }
template <class T, class... Ts>
void print(const T& t, const Ts&... ts) {
  pr(t);
  if (sizeof...(ts)) pr(" ");
  print(ts...);
}
}  // namespace output
using namespace output;
template <class T1, class T2>
istream& operator>>(istream& is, pair<T1, T2>& x) {
  is >> x.first >> x.second;
  return is;
}
template <class T1, class T2, class T3>
istream& operator>>(istream& is, tuple<T1, T2, T3>& x) {
  is >> get<0>(x) >> get<1>(x) >> get<2>(x);
  return is;
}
template <class T>
istream& operator>>(istream& is, vector<T>& v) {
  for (auto& x : v) is >> x;
  return is;
}
long long powa(long long base, long long exp) {
  long long res = 1;
  while (exp) {
    if (exp & 1) res = res * base % MOD;
    base = base * base % MOD;
    exp >>= 1;
  }
  return res;
}
bool mini(long long& a, long long b) { return b < a ? a = b, 1 : 0; }
bool maxi(long long& a, long long b) { return b > a ? a = b, 1 : 0; }
signed main() {
  ios::sync_with_stdio(0), cin.tie(0);
  long long n;
  cin >> n;
  vector<long long> res(n);
  long long i = 0, j = 1;
  while (i < n && j < n) {
    print('?', i + 1, j + 1);
    cout.flush();
    long long ij;
    cin >> ij;
    print('?', j + 1, i + 1);
    cout.flush();
    long long ji;
    cin >> ji;
    if (ij > ji) {
      res[i] = ij;
      while (i < n && res[i]) ++i;
      if (i == j) ++j;
    } else {
      res[j] = ji;
      while (j < n && res[j]) ++j;
    }
  }
  for (long long i = (0); i < (n); ++i) {
    if (res[i] == 0) {
      res[i] = n * (n + 1) / 2 - accumulate((res).begin(), (res).end(), 0LL);
      break;
    }
  }
  cout << "! ";
  for (long long i = (0); i < (n); ++i) cout << res[i] << ' ';
  print();
}