#include <bits/stdc++.h>
using namespace std;
bool comp(pair<long long, char> aa, pair<long long, char> bb) {
  if (aa.first != bb.first) return aa.first > bb.first;
  return aa.second < bb.second;
}
bool comp1(pair<long long, char> aa, pair<long long, char> bb) {
  if (aa.first != bb.first) return aa.first < bb.first;
  return aa.second > bb.second;
}
bool comp2(pair<long long, long long> aa, pair<long long, long long> bb) {
  if (aa.first != bb.first) return aa.first > bb.first;
  return aa.second < bb.second;
}
bool comp3(pair<long long, long long> aa, pair<long long, long long> bb) {
  if (aa.first != bb.first) return aa.first < bb.first;
  return aa.second > bb.second;
}
long long rup(long long ik, long long ikk) {
  if (ik % ikk == 0)
    return ik / ikk;
  else
    return (ik / ikk) + 1;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long modpow(long long x, long long n, long long m) {
  if (n == 0) return 1 % m;
  long long u = modpow(x, n / 2, m);
  u = (u * u) % m;
  if (n % 2 == 1) u = (u * x) % m;
  return u;
}
long long nm(long long a, long long b) { return (b + (a % b)) % b; }
struct hp {
  template <class T1, class T2>
  size_t operator()(const pair<T1, T2>& p) const {
    auto hash1 = hash<T1>{}(p.first);
    auto hash2 = hash<T2>{}(p.second);
    return hash1 ^ hash2;
  }
};
bool yo(string yy) {
  long long jh = yy.length() - 1;
  long long k = 0;
  long long f = 0;
  while (k < jh) {
    if (yy[k] != yy[jh]) {
      f++;
      break;
    }
    k++;
    jh--;
  }
  if (f == 0)
    return true;
  else
    return false;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    long long n;
    cin >> n;
    long long i;
    long long j;
    long long k;
    long long f = 0;
    long long v[105];
    for (i = 2; i <= n; i = i + 1) {
      cout << "? ";
      for (j = 1; j <= n - 1; j = j + 1) cout << i << " ";
      cout << 1 << "\n";
      cout.flush();
      cin >> k;
      if (k != 0) f++;
    }
    long long pp = f + 1;
    v[n] = pp;
    for (i = 1; i <= n; i = i + 1) {
      if (i == pp) continue;
      long long jh = abs(i - pp);
      if (i < pp) {
        cout << "? ";
        for (j = 1; j <= n - 1; j = j + 1) cout << 1 + jh << " ";
        cout << 1 << "\n";
        cout.flush();
        cin >> k;
        v[k] = pp - jh;
      } else {
        cout << "? ";
        for (j = 1; j <= n - 1; j = j + 1) cout << 1 << " ";
        cout << 1 + jh << "\n";
        cout.flush();
        cin >> k;
        v[k] = pp + jh;
      }
    }
    cout << "! ";
    for (i = 1; i <= n; i = i + 1) cout << v[i] << " ";
    cout << "\n";
    cout.flush();
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

It turns out that the meaning of life is a permutation p_1, p_2, …, p_n of the integers 1, 2, …, n (2 ≤ n ≤ 100). Omkar, having created all life, knows this permutation, and will allow you to figure it out using some queries.

A query consists of an array a_1, a_2, …, a_n of integers between 1 and n. a is not required to be a permutation. Omkar will first compute the pairwise sum of a and p, meaning that he will compute an array s where s_j = p_j + a_j for all j = 1, 2, …, n. Then, he will find the smallest index k such that s_k occurs more than once in s, and answer with k. If there is no such index k, then he will answer with 0.

You can perform at most 2n queries. Figure out the meaning of life p.

Interaction

Start the interaction by reading single integer n (2 ≤ n ≤ 100) — the length of the permutation p.

You can then make queries. A query consists of a single line "? \enspace a_1 \enspace a_2 \enspace … \enspace a_n" (1 ≤ a_j ≤ n).

The answer to each query will be a single integer k as described above (0 ≤ k ≤ n).

After making 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.



To output your answer, print a single line "! \enspace p_1 \enspace p_2 \enspace … \enspace p_n" then terminate.

You can make at most 2n queries. Outputting the answer does not count as a query.

Hack Format

To hack, first output a line containing n (2 ≤ n ≤ 100), then output another line containing the hidden permutation p_1, p_2, …, p_n of numbers from 1 to n.

Example

Input


5

2

0

1


Output



? 4 4 2 3 2

? 3 5 1 5 5

? 5 2 4 3 1

! 3 2 1 5 4

Note

In the sample, the hidden permutation p is [3, 2, 1, 5, 4]. Three queries were made.

The first query is a = [4, 4, 2, 3, 2]. This yields s = [3 + 4, 2 + 4, 1 + 2, 5 + 3, 4 + 2] = [7, 6, 3, 8, 6]. 6 is the only number that appears more than once, and it appears first at index 2, making the answer to the query 2.

The second query is a = [3, 5, 1, 5, 5]. This yields s = [3 + 3, 2 + 5, 1 + 1, 5 + 5, 4 + 5] = [6, 7, 2, 10, 9]. There are no numbers that appear more than once here, so the answer to the query is 0.

The third query is a = [5, 2, 4, 3, 1]. This yields s = [3 + 5, 2 + 2, 1 + 4, 5 + 3, 4 + 1] = [8, 4, 5, 8, 5]. 5 and 8 both occur more than once here. 5 first appears at index 3, while 8 first appears at index 1, and 1 < 3, making the answer to the query 1.

Note that the sample is only meant to provide an example of how the interaction works; it is not guaranteed that the above queries represent a correct strategy with which to determine the answer.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
bool comp(pair<long long, char> aa, pair<long long, char> bb) {
  if (aa.first != bb.first) return aa.first > bb.first;
  return aa.second < bb.second;
}
bool comp1(pair<long long, char> aa, pair<long long, char> bb) {
  if (aa.first != bb.first) return aa.first < bb.first;
  return aa.second > bb.second;
}
bool comp2(pair<long long, long long> aa, pair<long long, long long> bb) {
  if (aa.first != bb.first) return aa.first > bb.first;
  return aa.second < bb.second;
}
bool comp3(pair<long long, long long> aa, pair<long long, long long> bb) {
  if (aa.first != bb.first) return aa.first < bb.first;
  return aa.second > bb.second;
}
long long rup(long long ik, long long ikk) {
  if (ik % ikk == 0)
    return ik / ikk;
  else
    return (ik / ikk) + 1;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long modpow(long long x, long long n, long long m) {
  if (n == 0) return 1 % m;
  long long u = modpow(x, n / 2, m);
  u = (u * u) % m;
  if (n % 2 == 1) u = (u * x) % m;
  return u;
}
long long nm(long long a, long long b) { return (b + (a % b)) % b; }
struct hp {
  template <class T1, class T2>
  size_t operator()(const pair<T1, T2>& p) const {
    auto hash1 = hash<T1>{}(p.first);
    auto hash2 = hash<T2>{}(p.second);
    return hash1 ^ hash2;
  }
};
bool yo(string yy) {
  long long jh = yy.length() - 1;
  long long k = 0;
  long long f = 0;
  while (k < jh) {
    if (yy[k] != yy[jh]) {
      f++;
      break;
    }
    k++;
    jh--;
  }
  if (f == 0)
    return true;
  else
    return false;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    long long n;
    cin >> n;
    long long i;
    long long j;
    long long k;
    long long f = 0;
    long long v[105];
    for (i = 2; i <= n; i = i + 1) {
      cout << "? ";
      for (j = 1; j <= n - 1; j = j + 1) cout << i << " ";
      cout << 1 << "\n";
      cout.flush();
      cin >> k;
      if (k != 0) f++;
    }
    long long pp = f + 1;
    v[n] = pp;
    for (i = 1; i <= n; i = i + 1) {
      if (i == pp) continue;
      long long jh = abs(i - pp);
      if (i < pp) {
        cout << "? ";
        for (j = 1; j <= n - 1; j = j + 1) cout << 1 + jh << " ";
        cout << 1 << "\n";
        cout.flush();
        cin >> k;
        v[k] = pp - jh;
      } else {
        cout << "? ";
        for (j = 1; j <= n - 1; j = j + 1) cout << 1 << " ";
        cout << 1 + jh << "\n";
        cout.flush();
        cin >> k;
        v[k] = pp + jh;
      }
    }
    cout << "! ";
    for (i = 1; i <= n; i = i + 1) cout << v[i] << " ";
    cout << "\n";
    cout.flush();
  }
}

```