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>
#pragma GCC optimize("-Ofast", "-funroll-all-loops")
using namespace std;
long long int CeilIndex(std::vector<long long int>& v, long long int l,
                        long long int r, long long int key) {
  while (r - l > 1) {
    long long int m = l + (r - l) / 2;
    if (v[m] > key)
      r = m;
    else
      l = m;
  }
  return r;
}
long long int LongestIncreasingSubsequenceLength(
    std::vector<long long int>& v) {
  if (v.size() == 0) return 0;
  std::vector<long long int> tail(v.size(), 0);
  long long int length = 1;
  tail[0] = v[0];
  for (size_t i = 1; i < v.size(); i++) {
    if (v[i] < tail[0])
      tail[0] = v[i];
    else if (v[i] >= tail[length - 1])
      tail[length++] = v[i];
    else
      tail[CeilIndex(tail, -1, length - 1, v[i])] = v[i];
  }
  return length;
}
bool check(int m, vector<pair<int, int>> v) {
  int f = 0;
  for (int i = 0; i < v.size(); i++) {
    int x = v[i].first;
    int y = v[i].second;
    if (x > m && y > m) {
      f = 1;
      break;
    }
  }
  if (f == 1)
    return false;
  else
    return true;
}
long long int subsetSumToK(long long int arr[], long long int n,
                           long long int output[][50], long long int k) {
  if (n == 0) {
    if (k == 0) {
      output[0][0] = 0;
      return 1;
    } else {
      return 0;
    }
  }
  long long int output1[1000][50];
  long long int output2[1000][50];
  long long int size1 = subsetSumToK(arr + 1, n - 1, output1, k - arr[0]);
  long long int size2 = subsetSumToK(arr + 1, n - 1, output2, k);
  long long int i, j;
  for (i = 0; i < size1; i++) {
    output[i][0] = output1[i][0] + 1;
    output[i][1] = arr[0];
  }
  for (i = 0; i < size1; i++) {
    for (j = 1; j <= output1[i][0]; j++) {
      output[i][j + 1] = output1[i][j];
    }
  }
  for (i = 0; i < size2; i++) {
    for (j = 0; j <= output2[i][0]; j++) {
      output[i + size1][j] = output2[i][j];
    }
  }
  return size1 + size2;
}
void countSubsequences(long long int arr[], long long int n,
                       long long int output[][50], long long int k) {
  int size = subsetSumToK(arr, n, output, k);
  for (long long int i = 0; i < size; i++) {
    for (long long int j = 1; j <= output[i][0]; j++) {
      cout << output[i][j] << " ";
    }
    cout << endl;
  }
}
long long int subCount(long long int arr[], long long int n, long long int k) {
  long long int mod[k];
  memset(mod, 0, sizeof(mod));
  long long int cumSum = 0;
  for (long long int i = 0; i < n; i++) {
    cumSum += arr[i];
    mod[((cumSum % k) + k) % k]++;
  }
  long long int result = 0;
  for (long long int i = 0; i < k; i++)
    if (mod[i] > 1) result += (mod[i] * (mod[i] - 1)) / 2;
  result += mod[0];
  return result;
}
long long int fact(long long int n);
long long int nCr(long long int n, long long int r) {
  return fact(n) / (fact(r) * fact(n - r));
}
long long int fact(long long int n) {
  long long int res = 1;
  for (long long int i = 2; i <= n; i++) res = res * i;
  return res;
}
long long int subsetPairNotDivisibleByK(long long int arr[], long long int N,
                                        long long int K) {
  long long int f[K];
  memset(f, 0, sizeof(f));
  for (long long int i = 0; i < N; i++) f[arr[i] % K]++;
  if (K % 2 == 0) f[K / 2] = min(f[K / 2], 1LL);
  long long int res = min(f[0], 1LL);
  for (long long int i = 1; i <= K / 2; i++) res += max(f[i], f[K - i]);
  return res;
}
bool isPerfectSquare(long double x) {
  if (x >= 0) {
    long double sr = sqrt(x);
    return (sr * sr == x);
  }
  return false;
}
vector<long long int> v[1000001];
long long int dis[50][50];
long long int deg[1000001];
long long int ask(long long int x, long long int y) {
  cout << '?' << " " << x << " " << y << '\n';
  cout.flush();
  long long int w;
  cin >> w;
  return w;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n;
  cin >> n;
  long long int x12 = ask(1, n);
  long long int x13 = ask(1, n - 1);
  long long int x23 = ask(n, n - 1);
  long long int a[n + 1];
  a[n] = (x12 - x13 + x23) / 2;
  a[1] = x12 - a[n];
  a[n - 1] = x13 - a[1];
  for (long long int i = n - 2; i >= 2; i--) {
    long long int y = ask(i, i + 1);
    a[i] = y - a[i + 1];
  }
  cout << '!' << " ";
  for (long long int i = 1; i <= n; i++) cout << a[i] << " ";
  cout.flush();
  return 0;
}