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
#include <bits/stdc++.h>
using namespace std;
int bPow(int v, int u, int mod) {
  if (!u) return 1;
  int res = bPow(v, u / 2, mod);
  res = (res * 1ll * res) % mod;
  if (u & 1) res = (res * 1ll * v) % mod;
  return res;
}
void faster() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  if (0) {
    freopen(
        "a"
        ".in",
        "r", stdin);
    freopen(
        "a"
        ".out",
        "w", stdout);
  }
}
vector<int> prime;
bool notprime[32000];
void resh() {
  int n = 32000 - 10;
  for (int i = 2; i < n; i++)
    if (!notprime[i]) {
      prime.push_back(i);
      for (int j = i; j < n; j += i) notprime[j] = 1;
    }
}
int is_comp(long long p) {
  for (int i = 0; prime[i] * 1ll * prime[i] <= p; i++) {
    if (p % prime[i] == 0) return prime[i];
  }
  return 0;
}
bool is_prime(long long p) {
  for (int i = 0; prime[i] * 1ll * prime[i] <= p; i++) {
    if (p % prime[i] == 0) return 0;
  }
  return 1;
}
long long gcd(long long a, long long b) {
  if (a > b) swap(a, b);
  long long temp;
  while (a) {
    temp = a;
    a = b % a;
    b = temp;
  }
  return b;
}
int n, p[11111], k;
void ask(int x, int y) {
  cout << "? " << x + 1 << ' ' << y + 1 << endl;
  cout.flush();
  cin >> k;
}
void ans() {
  cout << "! ";
  for (int i = 0; i < n; i++) cout << p[i] << ' ';
  cout << endl;
  cout.flush();
}
bool great(int x, int y) {
  int a, b;
  ask(x, y);
  a = k;
  ask(y, x);
  b = k;
  if (a < b)
    p[y] = b;
  else
    p[x] = a;
  return a < b;
}
int maxn() {
  int x = 0;
  for (int i = 1; i < n; i++) {
    if (great(i, x)) x = i;
  }
  return x;
}
void solve() {
  cin >> n;
  if (n == 1) {
    p[0] = 1;
    ans();
    return;
  }
  int x = maxn();
  p[x] = n;
  ans();
}
int main() {
  faster();
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}
