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;
template <class A, class B>
ostream &operator<<(ostream &out, const pair<A, B> &a);
template <class A>
ostream &operator<<(ostream &out, const vector<A> &v);
void swap(int &a, int &b);
vector<int> sieveOfEratosthenes(int n);
bool isPrime(int number);
vector<int> returnAllFactors(int n);
int totalNumberOfDigits(int A);
int extractIthDigit(int A, int i);
int reverseDigits(int A);
vector<int> allPrimeFactors(int A);
int gcdEuclidean(int A, int B);
int lcm(int A, int B);
long long lb(long long target, long long l, long long h);
long long ub(long long target, long long l, long long h);
long long f(long long x) {
  long long answer;
  return answer;
}
int ask(int i, int j) {
  cout << "? " << i + 1 << " " << j + 1 << endl;
  int x;
  cin >> x;
  return x;
}
void solve() {
  int t, n, m, l, r, w, x, y, z, k, temp, answer, mini, maxi, miniIndex,
      maxiIndex, counter, sum;
  bool cond, parity;
  string str, str1;
  t = 1;
  while (t--) {
    cin >> n;
    int mx = 0;
    vector<int> arr(n, -1);
    for (int i = 1; i < n; i++) {
      int a = ask(mx, i);
      int b = ask(i, mx);
      if (a > b) {
        arr[mx] = a;
        mx = i;
      } else {
        arr[i] = b;
      }
    }
    arr[mx] = n;
    cout << "! ";
    for (int i = 0; i < int(n); i++) {
      cout << arr[i] << " ";
    }
    cout << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}
template <class A, class B>
ostream &operator<<(ostream &out, const pair<A, B> &a) {
  return out << "(" << a.first << ", " << a.second << ")";
}
template <class A>
ostream &operator<<(ostream &out, const vector<A> &v) {
  out << "[";
  for (int i = 0; i < int(int(v.size())); i++) {
    if (i) out << ", ";
    out << v[i];
  }
  return out << "]";
}
void swap(int &a, int &b) {
  int temp;
  temp = a;
  a = b;
  b = temp;
}
vector<int> sieveOfEratosthenes(int n) {
  vector<int> primes(n + 1, 1);
  vector<int> resultVector;
  primes[0] = 0;
  primes[1] = 0;
  for (int i = 2; i <= sqrt(n); i++) {
    if (primes[i] == 1) {
      for (int j = i * i; j <= n; j += i) {
        primes[j] = 0;
      }
    }
  }
  for (int i = 2; i <= n; i++) {
    if (primes[i] == 1) {
      resultVector.push_back(i);
    }
  }
  return resultVector;
}
bool isPrime(int number) {
  bool cond = true;
  if (number <= 1) {
    cond = false;
  } else if (number == 2 || number == 3) {
    cond = true;
  } else {
    for (int i = 2; i <= sqrt(number); i++) {
      if (number % i == 0) {
        cond = false;
        break;
      } else {
        continue;
      }
    }
  }
  return cond;
}
vector<int> returnAllFactors(int n) {
  vector<int> result;
  for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      result.push_back(i);
      if (i != sqrt(n)) {
        result.push_back(n / i);
      }
    }
  }
  sort((result).begin(), (result).end());
  return result;
}
int totalNumberOfDigits(int A) {
  int result = 0;
  while (A > 0) {
    A = A / 10;
    result++;
  }
  return result;
}
int extractIthDigit(int A, int i) {
  int result = (int(A / pow(10, i)) % 10);
  return result;
}
int reverseDigits(int A) {
  bool negative_number = false;
  if (A < 0) {
    negative_number = true;
    A = A * -1;
  }
  long long reverse = 0;
  int prev_reverse = 0;
  while (A > 0) {
    int current_digit = A % 10;
    reverse = reverse * 10 + current_digit;
    if ((reverse - current_digit) / 10 != prev_reverse) {
      return 0;
    }
    prev_reverse = reverse;
    A = A / 10;
  }
  if (negative_number && -reverse < INT32_MIN)
    return 0;
  else if (reverse > INT32_MAX)
    return 0;
  else
    return negative_number ? -reverse : reverse;
}
int gcdEuclidean(int A, int B) {
  if (B == 0) return A;
  return gcdEuclidean(B, A % B);
}
int lcm(int A, int B) { return (A * B) / gcdEuclidean(A, B); }
long long lb(long long target, long long l, long long h) {
  long long mid, fmid;
  long long answer = INT64_MAX;
  while (l <= h) {
    mid = (l + h) / 2;
    fmid = f(mid);
    if (fmid >= target) {
      h = mid - 1;
      answer = min(answer, mid);
    } else {
      l = mid + 1;
    }
  }
  return answer;
}
long long ub(long long target, long long l, long long h) {
  long long mid, fmid;
  long long answer = INT64_MAX;
  while (l <= h) {
    mid = (l + h) / 2;
    fmid = f(mid);
    if (fmid > target) {
      h = mid - 1;
      answer = min(answer, mid);
    } else {
      l = mid + 1;
    }
  }
  return answer;
}