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")
using namespace std;
long long TESTCASE_NUM = 1;
long long TESTCASE_CUR = 0;
template <typename T>
inline void _read(T &x) {
  cin >> x;
}
inline void _read(int &x) { scanf("%d", &x); }
inline void _read(long long &x) { scanf("%lld", &x); }
inline void _read(float &x) { scanf("%f", &x); }
inline void _read(double &x) { scanf("%lf", &x); }
inline void _read(char &x) { x = getchar(); }
inline void _read(char *x) { scanf("%s", x); }
inline void read() {}
template <typename T, typename... U>
inline void read(T &head, U &...tail) {
  _read(head);
  read(tail...);
}
template <typename T>
inline void _write(const T &x) {
  cout << x;
}
inline void _write(const int &x) { printf("%d", x); }
inline void _write(const long long &x) { printf("%lld", x); }
inline void _write(const float &x) { printf("%.6f", x); }
inline void _write(const double &x) { printf("%.8lf", x); }
inline void _write(const char &x) { putchar(x); }
inline void _write(const char *x) { printf("%s", x); }
inline void write() {}
template <typename T, typename... U>
inline void write(const T &head, const U &...tail) {
  _write(head);
  write(tail...);
}
inline void debug() {}
template <typename T, typename... U>
inline void debug(const T &head, const U &...tail) {}
template <typename T>
inline bool isprime(T p) {
  if (p == 2 || p == 3) return 1;
  if (p < 2 || (p % 6 != 1 && p % 6 != 5)) return 0;
  for (T i = 5; i * i <= p; i = i + 6)
    if (p % i == 0 || p % (i + 2) == 0) return 0;
  return 1;
}
template <typename T1, typename T2, typename T3>
inline T1 qpow(T1 a, T2 b, T3 m) {
  T1 ans = 1, base = a;
  while (b) {
    if (b & 1) ans = ans * base % m;
    base = base * base % m;
    b >>= 1;
  }
  return ans;
}
template <typename T1, typename T2>
inline T1 qpow(T1 a, T2 b) {
  T1 ans = 1, base = a;
  while (b) {
    if (b & 1) ans = ans * base;
    base = base * base;
    b >>= 1;
  }
  return ans;
}
template <typename T1, typename T2>
inline T1 inv(T1 i, T2 m) {
  return qpow(i, m - 2, m);
}
template <typename T1, typename T2>
inline T1 max(T1 a, T2 b) {
  return a < b ? b : a;
}
template <typename T1, typename T2>
inline T1 min(T1 a, T2 b) {
  return a < b ? a : b;
}
void solve();
void init();
signed main() {
  init();
  while (TESTCASE_CUR++ < TESTCASE_NUM) {
    solve();
  }
  return 0;
}
const int MAXN = 1e5 + 10;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
void init() {}
long long ask(long long x, long long y) {
  write("? ", x, ' ', y, '\n');
  fflush(stdout);
  long long res;
  read(res);
  return res;
}
void solve() {
  long long n;
  read(n);
  long long ans[n + 10];
  int maxi = 1;
  for (int i = 2; i <= n; i++) {
    long long a = ask(maxi, i);
    long long b = ask(i, maxi);
    if (a > b) {
      ans[maxi] = a;
      maxi = i;
    } else {
      ans[i] = b;
    }
  }
  ans[maxi] = n;
  write("! ");
  for (int i = 1; i <= n; i++) {
    write(ans[i], ' ');
  }
  write('\n');
}