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;
  template <typename A, typename B>
  ostream &operator<<(ostream &os, const pair<A, B> &p) {
    return os << '(' << p.first << ", " << p.second << ')';
  }
  template <typename T_container, typename T = typename enable_if<
                                      !is_same<T_container, string>::value,
                                      typename T_container::value_type>::type>
  ostream &operator<<(ostream &os, const T_container &v) {
    os << '{';
    string sep;
    for (const T &x : v) os << sep << x, sep = ", ";
    return os << '}';
  }
  void dbg_out() { cout << endl; }
  template <typename Head, typename... Tail>
  void dbg_out(Head H, Tail... T) {
    cout << ' ' << H;
    dbg_out(T...);
  }
  int32_t main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    long long tt;
    cin >> tt;
    while (tt--) {
      long long n;
      cin >> n;
      vector<long long> a(n);
      for (long long i = 0; i < n; i++) cin >> a[i];
      long long sum = accumulate((a).begin(), (a).end(), 0ll);
      long long tot = (n * (n + 1)) / 2;
      if (sum % tot || sum / tot < n) {
        cout << "NO\n";
        continue;
      }
      sum /= tot;
      vector<long long> ans(n);
      long long bad = 0;
      for (long long i = 0; i < n; i++) {
        long long diff = a[(i + 1) % n] - a[i];
        if ((sum - diff) % n) bad = 1;
        ans[(i + 1) % n] = (sum - diff) / n;
      };
      long long s = 0;
      for (auto x : ans) bad |= (x <= 0), s += x;
      if (bad || sum != s)
        cout << "NO\n";
      else {
        cout << "YES\n";
        for (auto x : ans) cout << x << " ";
        cout << "\n";
      }
    }
  }
  