Problem: This is an interactive problem. You should use flush operation after each printed line. For example, in C++ you should use fflush(stdout), in Java you should use System.out.flush(), and in Pascal — flush(output).

In this problem you should guess an array a which is unknown for you. The only information you have initially is the length n of the array a.

The only allowed action is to ask the sum of two elements by their indices. Formally, you can print two indices i and j (the indices should be distinct). Then your program should read the response: the single integer equals to ai + aj.

It is easy to prove that it is always possible to guess the array using at most n requests.

Write a program that will guess the array a by making at most n requests.

Interaction

In each test your program should guess a single array.

The input starts with a line containing integer n (3 ≤ n ≤ 5000) — the length of the array. Your program should read it at first.

After that your program should print to the standard output the requests about the sum of two elements or inform that the array is guessed.

  * In case your program is making a request to ask the sum of two elements, it should print line in the format "? i j" (i and j are distinct integers between 1 and n), where i and j are indices in the array a.
  * In case your program informs that the array is guessed, it should print line in the format "! a1 a2 ... an" (it is guaranteed that all ai are positive integers not exceeding 105), where ai is the i-th element of the array a. 



The response on a request is a single integer equal to ai + aj, printed on a separate line.

Your program can do at most n requests. Note that the final line «! a1 a2 ... an» is not counted as a request.

Do not forget about flush operation after each printed line.

After you program prints the guessed array, it should terminate normally.

Example

Input

5
 
9
 
7
 
9
 
11
 
6
 

Output

 
? 1 5
 
? 2 3
 
? 4 1
 
? 5 2
 
? 3 4
 
! 4 6 1 5 5

Note

The format of a test to make a hack is:

  * The first line contains an integer number n (3 ≤ n ≤ 5000) — the length of the array.
  * The second line contains n numbers a1, a2, ..., an (1 ≤ ai ≤ 105) — the elements of the array to guess. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using str = string;
using vli = vector<int>;
using vll = vector<long long int>;
using sli = set<int>;
using sll = set<long long int>;
using pli = pair<int, int>;
using pll = pair<long long int, long long int>;
using vpi = vector<pair<int, int>>;
using vpl = vector<pair<long long int, long long int>>;
using mpl = map<long long int, long long int>;
using qll = queue<long long int>;
using stl = stack<long long int>;
const double eps = 1e-9;
const int INF = 2000000000;
const long long int infLL = 9000000000000000000;
const long long int MOD = 1e9 + 7;
const double PI = 3.141592653589793238462643383279;
long long o_O = 0;
long long ceil(long long int x, long long int y) {
  return x / y + (x % y != 0);
}
int dx[] = {0, 0, +1, -1, +1, +1, -1, -1};
int dy[] = {+1, -1, 0, 0, +1, -1, +1, -1};
const int mX = 3e5 + 123;
long long int n, p, a[mX];
void case_i() {
  cin >> n;
  cout << '?' << " " << 2 << " " << 3 << '\n';
  ;
  cout << flush;
  cin >> p;
  for (long long int i = 2; i <= n; i++) {
    cout << '?' << " " << 1 << " " << i << '\n';
    ;
    cout << flush;
    cin >> a[i];
  }
  p = (a[2] + a[3] - p) / 2;
  cout << '!' << " " << p << " ";
  for (long long int i = 2; i <= n; i++) {
    cout << a[i] - p << " ";
  }
  std ::cout << '\n';
}
int32_t main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  case_i();
  return 0;
}
long long int _pow(long long int p, long long int q) {
  if (q == 0) return 1;
  return p * _pow(p, q - 1);
}
long long int pow_mod(long long int x, long long int n, long long int m,
                      long long int k = 1) {
  x %= m;
  while (n > 0) {
    if (n & 1) k = (k * x) % m;
    x = (x * x) % m;
    n >>= 1;
  }
  return k;
}
long long int mod_inverse(long long int n) { return pow_mod(n, MOD - 2, MOD); }
long long int factorial(long long int f) {
  if (f < 2) return 1;
  return f * factorial(f - 1);
}
bool is_prime(long long int n) {
  if (n <= 3 and n != 1) return 1;
  if (n % 2 == 0 or n % 3 == 0) return 0;
  for (long long int i = 5; i * i <= n; i += 6)
    if (n % i == 0 or n % (i + 2) == 0) return 0;
  return 1;
}
void sieve_prime_generator(long long int n, bool vis[],
                           vector<long long int>& prime) {
  prime.push_back(2);
  for (long long int i = 3; i <= n;
       (!vis[i]) ? prime.push_back(i), i += 2 : i += 2)
    if (!vis[i])
      for (long long int j = i * i; j <= n; vis[j] = true, j += (2 * i))
        ;
}
void prime_factorization(long long int n,
                         vector<pair<long long int, long long int>>& prime_fact,
                         vector<long long int>& prime, long long int k = 0) {
  prime_fact.clear();
  for (auto i : prime) {
    if (i * i > n) break;
    while (n % i == 0) n /= i, k++;
    if (k) prime_fact.push_back({i, k}), k = 0;
  }
  if (n > 1) prime_fact.push_back({n, 1});
}
void total_divisors(long long int a[], long long int n) {
  for (long long int i = 1; i * i <= n; i++)
    for (long long int j = i * i; j < n; j += i)
      if (j == i * i)
        a[j]++;
      else
        a[j] += 2;
}
void sum_of_divisors(long long int a[], long long int n) {
  for (long long int i = 1; i * i <= n; i++)
    for (long long int j = i * i; j < n; j += i)
      if (j == i * i)
        a[j] += i;
      else
        a[j] += i + j / i;
}
void dfs(long long int i, bool vis[], vector<long long int> v[]) {
  vis[i] = true;
  for (auto j : v[i])
    if (!vis[j]) dfs(j, vis, v);
}
void bfs(long long int i, vector<long long int> v[], long long int lev[]) {
  queue<long long int> q;
  q.push(i);
  lev[i] = 0;
  while (!q.empty()) {
    long long int u = q.front();
    q.pop();
    for (auto j : v[u])
      if (lev[j] == -1 or lev[j] > lev[u] + 1) {
        lev[j] = lev[u] + 1;
        q.push(j);
      }
  }
}