Problem: This is an interactive problem.

Ayush devised yet another scheme to set the password of his lock. The lock has n slots where each slot can hold any non-negative integer. The password P is a sequence of n integers, i-th element of which goes into the i-th slot of the lock.

To set the password, Ayush comes up with an array A of n integers each in the range [0, 2^{63}-1]. He then sets the i-th element of P as the [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR) of all integers in the array except A_i.

You need to guess the password. To make a query, you can choose a non-empty subset of indices of the array and ask the bitwise OR all elements of the array with index in this subset. You can ask no more than 13 queries.

Input

The first line of input contains one integer n (2 ≤ n ≤ 1000) — the number of slots in the lock.

Interaction

To ask a query print a single line: 

  * In the beginning print "? c " (without quotes) where c (1 ≤ c ≤ n) denotes the size of the subset of indices being queried, followed by c distinct space-separated integers in the range [1, n]. 



For each query, you will receive an integer x — the bitwise OR of values in the array among all the indices queried. If the subset of indices queried is invalid or you exceeded the number of queries then you will get x = -1. In this case, you should terminate the program immediately.

When you have guessed the password, print a single line "! " (without quotes), followed by n space-separated integers — the password sequence.

Guessing the password does not count towards the number of queries asked.

The interactor is not adaptive. The array A does not change with queries.

After printing a query do not forget to output the end of the 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 the documentation for other languages.



Hacks

To hack the solution, use the following test format:

On the first line print a single integer n (2 ≤ n ≤ 1000) — the number of slots in the lock. The next line should contain n space-separated integers in the range [0, 2^{63} - 1] — the array A.

Example

Input


3

1

2

4



Output


? 1 1

? 1 2

? 1 3

! 6 5 3

Note

The array A in the example is \{{1, 2, 4\}}. The first element of the password is bitwise OR of A_2 and A_3, the second element is bitwise OR of A_1 and A_3 and the third element is bitwise OR of A_1 and A_2. Hence the password sequence is \{{6, 5, 3\}}.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T &x) {
  static char _c;
  static bool _f;
  x = 0;
  _f = 0;
  _c = getchar();
  while (_c < '0' || '9' < _c) {
    if (_c == '-') _f = true;
    _c = getchar();
  }
  while ('0' <= _c && _c <= '9') {
    x = (x << 1) + (x << 3) + (_c & 15);
    _c = getchar();
  }
  if (_f) x = -x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &...args) {
  read(x);
  read(args...);
}
template <typename T>
inline void Min(T &x, T y) {
  if (y < x) x = y;
}
template <typename T>
inline void Max(T &x, T y) {
  if (x < y) x = y;
}
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const double pi = (double)acos(-1.0);
const double eps = (double)1e-8;
const int e5 = (int)1e5 + 5;
const int MOD = (int)1e9 + 7;
template <typename T0, typename T1>
inline void depair(T0 &x, T1 &y, pair<T0, T1> &p) {
  x = p.first, y = p.second;
}
inline int sig(double x) { return x < -eps ? -1 : eps < x; }
long long fp(long long a, long long n, long long mod = MOD) {
  if (n < 0) a = fp(a, mod - 2, mod), n = -n;
  long long res = 1;
  for (; n; n >>= 1, a = a * a % mod)
    if (n & 1) res = res * a % mod;
  return res;
}
struct Mint {
  int x;
  Mint() : x(0) {}
  Mint(int _x) : x(_x) {
    if (x >= MOD)
      x %= MOD;
    else if (x < 0)
      x = (x % MOD + MOD) % MOD;
  }
  Mint operator-() const { return Mint(MOD - x); }
  Mint operator+(const Mint &rhs) const {
    return Mint(x + rhs.x >= MOD ? x - MOD + rhs.x : x + rhs.x);
  }
  Mint operator-(const Mint &rhs) const {
    return Mint(x - rhs.x < 0 ? x - rhs.x + MOD : x - rhs.x);
  }
  Mint operator*(const Mint &rhs) const {
    return Mint((long long)x * rhs.x % MOD);
  }
  Mint operator/(const Mint &rhs) const {
    return Mint((long long)x * fp(rhs.x, -1) % MOD);
  }
  Mint &operator+=(const Mint &rhs) {
    x += rhs.x;
    if (x >= MOD) x -= MOD;
    return *this;
  }
  Mint &operator-=(const Mint &rhs) {
    x -= rhs.x;
    if (x < 0) x += MOD;
    return *this;
  }
  Mint &operator*=(const Mint &rhs) {
    x = ((long long)x * rhs.x) % MOD;
    return *this;
  }
  Mint &operator/=(const Mint &rhs) {
    x = (long long)x * fp(rhs.x, -1) % MOD;
    return *this;
  }
  bool operator==(const Mint &rhs) const { return x == rhs.x; }
  bool operator!=(const Mint &rhs) const { return x != rhs.x; }
  friend ostream &operator<<(ostream &out, const Mint &rhs) {
    return out << rhs.x;
  }
  friend istream &operator>>(istream &in, Mint &rhs) { return in >> rhs.x; }
  Mint rev() { return Mint(fp(x, -1)); }
  Mint pow(int k) { return Mint(fp(x, k)); }
  Mint pow(const Mint &t) { return Mint(fp(x, t.x)); }
};
const int maxn = (int)1e6 + 20;
const int maxm = (int)2e6 + 20;
void work() {
  int n;
  cin >> n;
  vector<int> mask(n);
  for (int i = 0, j = 0; i < n; i++) {
    auto popcount = [](int x) {
      int cnt = 0;
      while (x) {
        cnt++;
        x -= ((x) & -(x));
      }
      return cnt;
    };
    while (popcount(j) != 6) j++;
    mask[i] = j++;
  }
  vector<long long> v(13);
  for (int i = 0; i < 13; i++) {
    vector<int> s;
    for (int j = 0; j < n; j++)
      if (mask[j] >> i & 1) s.push_back(j);
    if (s.size() == 0) {
      v[i] = 0;
    } else {
      printf("? %d ", s.size());
      for (int k = 0; k < s.size(); k++)
        printf("%d%c", s[k] + 1, " \n"[k == s.size() - 1]);
      fflush(stdout);
      read(v[i]);
    }
  }
  vector<long long> ans(n, 0);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 13; j++)
      if (~mask[i] >> j & 1) ans[i] |= v[j];
  }
  printf("! ");
  for (int i = 0; i < n; i++) printf("%lld%c", ans[i], " \n"[i == n - 1]);
  fflush(stdout);
}
int main() {
  int tc = 1;
  for (int ca = 1; ca <= tc; ca++) {
    work();
  }
  return 0;
}