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;
  const int inf = 1e9 + 1;
  const long long infll = 1e18 + 1;
  const int mod = 998244353;
  mt19937_64 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
  template <typename T>
  bool chmin(T &a, T b) {
    if (a > b) {
      a = b;
      return true;
    }
    return false;
  }
  template <typename T>
  bool chmax(T &a, T b) {
    if (a < b) {
      a = b;
      return true;
    }
    return false;
  }
  struct smth {
    int x;
  };
  void solve() {
    int n;
    cin >> n;
    long long b[n];
    long long s = 0;
    for (int i = 0; i < n; i++) {
      cin >> b[i];
      s += b[i];
    }
    long long bg = 1ll * n * (n + 1) / 2;
    if (s % bg != 0) {
      cout << "NO\n";
      return;
    }
    s /= bg;
    vector<long long> a;
    for (int i = 0; i < n; i++) {
      long long cur;
      if (i == 0) {
        cur = b[n - 1] - b[0] + s;
      } else {
        cur = b[i - 1] - b[i] + s;
      }
      if (cur <= 0 || cur % n != 0) {
        cout << "NO\n";
        return;
      }
      a.push_back(cur / n);
    }
    cout << "YES\n";
    for (long long i : a) {
      cout << i << ' ';
    }
    cout << '\n';
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int t;
    t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  