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;
const int maxn = 1e5 + 200;
const int inf = 0x3f3f3f3f;
const double eps = 1e-7;
const double pi = acos(-1.0);
const int mod = 1e9 + 7;
inline int lowbit(int x) { return x & (-x); }
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
void ex_gcd(long long a, long long b, long long &d, long long &x,
            long long &y) {
  if (!b) {
    d = a, x = 1, y = 0;
  } else {
    ex_gcd(b, a % b, d, y, x);
    y -= x * (a / b);
  }
}
inline long long qpow(long long a, long long b, long long MOD = mod) {
  long long res = 1;
  a %= MOD;
  while (b > 0) {
    if (b & 1) res = res * a % MOD;
    a = a * a % MOD;
    b >>= 1;
  }
  return res;
}
inline long long inv(long long x, long long p) { return qpow(x, p - 2, p); }
inline long long Jos(long long n, long long k, long long s = 1) {
  long long res = 0;
  for (int i = 1; i <= n + 1; ++i) res = (res + k) % i;
  return (res + s) % n;
}
inline long long read() {
  long long f = 1;
  long long x = 0;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return x * f;
}
int dir[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
long long a[maxn];
long long vis[maxn];
void solve() {
  long long n = read();
  int p = 1, q = n;
  int cnt = 0;
  while (cnt <= n - 1 && p < q) {
    printf("? %d %d\n", p, q);
    fflush(stdout);
    long long r1 = read();
    printf("? %d %d\n", q, p);
    fflush(stdout);
    long long r2 = read();
    if (r1 > r2)
      a[p] = r1, p++, vis[r1] = 1;
    else
      a[q] = r2, q--, vis[r2] = 1;
    cnt++;
  }
  long long sum = (1 + n) * n / 2;
  long long all = 0;
  for (int i = 1; i <= n; ++i) all += a[i];
  for (int i = 1; i <= n; ++i)
    if (a[i] == 0) a[i] = sum - all;
  printf("! ");
  for (int i = 1; i <= n; ++i) printf("%lld%c", a[i], i == n ? '\n' : ' ');
  fflush(stdout);
}
int main() {
  solve();
  return 0;
}
