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;
template <class T>
void read(T &x) {
  T res = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    res = (res << 3) + (res << 1) + c - '0';
    c = getchar();
  }
  x = res * f;
}
int gcd(int a, int b) { return (b == 0) ? a : gcd(b, a % b); }
const int inf = INT_MAX;
const int maxn = 1e5 + 10;
const int mod = 1e9 + 7;
int n, a[maxn], vis[maxn], b, c, d[maxn];
int main() {
  int T = 1;
  while (T--) {
    memset(d, 0, sizeof(d)), memset(vis, 0, sizeof(vis));
    read(n);
    if (n == 1) {
      cout << "!"
           << " " << 1 << endl;
      break;
    }
    int l = 1, r = 2;
    for (int i = 1; i <= n - 1; i++) {
      printf("? %d %d\n", l, r);
      fflush(stdout);
      scanf("%d", &b);
      printf("? %d %d\n", r, l);
      fflush(stdout);
      scanf("%d", &c);
      if (b > c)
        d[l] = b, l = r, r = r + 1, vis[b] = 1;
      else
        d[r] = c, r = r + 1, vis[c] = 1;
    }
    int p, x;
    for (int i = 1; i <= n; i++) {
      if (d[i] == 0) p = i;
      if (vis[i] == 0) x = i;
    }
    d[p] = x;
    cout << "!"
         << " ";
    for (int i = 1; i <= n; i++) {
      cout << d[i] << " ";
    }
    cout << endl;
  }
  return 0;
}
