description: n towns are arranged in a circle sequentially. The towns are numbered from 1 to n in clockwise order. In the i-th town, there lives a singer with a repertoire of a_i minutes for each i ∈ [1, n].
  
  Each singer visited all n towns in clockwise order, starting with the town he lives in, and gave exactly one concert in each town. In addition, in each town, the i-th singer got inspired and came up with a song that lasts a_i minutes. The song was added to his repertoire so that he could perform it in the rest of the cities.
  
  Hence, for the i-th singer, the concert in the i-th town will last a_i minutes, in the (i + 1)-th town the concert will last 2 ⋅ a_i minutes, ..., in the ((i + k) mod n + 1)-th town the duration of the concert will be (k + 2) ⋅ a_i, ..., in the town ((i + n - 2) mod n + 1) — n ⋅ a_i minutes.
  
  You are given an array of b integer numbers, where b_i is the total duration of concerts in the i-th town. Reconstruct any correct sequence of positive integers a or say that it is impossible.
  
  Input
  
  The first line contains one integer t (1 ≤ t ≤ 10^3) — the number of test cases. Then the test cases follow.
  
  Each test case consists of two lines. The first line contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^4) — the number of cities. The second line contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^{9}) — the total duration of concerts in i-th city.
  
  The sum of n over all test cases does not exceed 2 ⋅ 10^5.
  
  Output
  
  For each test case, print the answer as follows:
  
  If there is no suitable sequence a, print NO. Otherwise, on the first line print YES, on the next line print the sequence a_1, a_2, ..., a_n of n integers, where a_i (1 ≤ a_i ≤ 10^{9}) is the initial duration of repertoire of the i-th singer. If there are multiple answers, print any of them.
  
  Example
  
  Input
  
  
  4
  3
  12 16 14
  1
  1
  3
  1 2 3
  6
  81 75 75 93 93 87
  
  
  Output
  
  
  YES
  3 1 3 
  YES
  1 
  NO
  YES
  5 5 4 1 4 5 
  
  Note
  
  Let's consider the 1-st test case of the example:
  
    1. the 1-st singer in the 1-st city will give a concert for 3 minutes, in the 2-nd — for 6 minutes, in the 3-rd — for 9 minutes; 
    2. the 2-nd singer in the 1-st city will give a concert for 3 minutes, in the 2-nd — for 1 minute, in the 3-rd - for 2 minutes; 
    3. the 3-rd singer in the 1-st city will give a concert for 6 minutes, in the 2-nd — for 9 minutes, in the 3-rd — for 3 minutes. 
solution: #include <bits/stdc++.h>
  using namespace std;
  constexpr long long int mod = 1e9 + 7;
  auto seed = chrono::high_resolution_clock::now().time_since_epoch().count();
  std::mt19937 mt(seed);
  int myrand(int mod) { return (int)((((double)rand() / RAND_MAX) * (mod))); }
  template <typename T, typename U>
  void operator%=(T &a, const U &m) {
    if (abs(a) >= m) a %= m;
    if (a < 0) a += m;
  }
  template <typename T, typename U>
  T operator%(T &a, const U &m) {
    T b = a;
    b %= m;
    return b;
  }
  template <typename T>
  ostream &operator<<(ostream &out, const vector<T> &a) {
    for (int i = int(0); i < int((int)a.size()); i++) out << a[i] << " ";
    out << "\n";
    return out;
  }
  template <typename T>
  istream &operator>>(istream &in, vector<T> &a) {
    for (int i = int(0); i < int((int)a.size()); i++) in >> a[i];
    return in;
  }
  template <typename T, typename U>
  T pow2(const T &x, const U &y) {
    T res = 1, X = x;
    U Y = y;
    while (Y > 0) {
      if (Y & 1) res *= X;
      Y >>= 1;
      X *= X;
      X %= mod;
      res %= mod;
    }
    return res;
  }
  pair<long long int, long long int> operator-(
      pair<long long int, long long int> &a,
      pair<long long int, long long int> &b) {
    return {a.first - b.first, a.second - b.second};
  }
  pair<long long int, long long int> operator+(
      pair<long long int, long long int> &a,
      pair<long long int, long long int> &b) {
    return {a.first + b.first, a.second + b.second};
  }
  void solve() {
    int n;
    cin >> n;
    vector<long long int> a(n);
    cin >> a;
    long long int N = n * (n + 1) / 2;
    long long int sum = accumulate(a.begin(), a.end(), 0LL);
    if (sum % N != 0) {
      cout << "NO\n";
      return;
    }
    sum /= N;
    vector<long long int> res(n);
    for (int i = 0; i < n; i++) {
      int j = (i + 1) % n;
      long long int c = a[j] - a[i];
      res[j] = sum - c;
      if (res[j] <= 0 || res[j] % n != 0) {
        cout << "NO\n";
        return;
      } else {
        res[j] /= n;
      }
    }
    cout << "YES\n";
    cout << res;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t = 1;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  