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 long long maxn = 1e6 + 50;
  template <typename A, typename B>
  istream& operator>>(istream& o, pair<A, B>& a) {
    return o >> a.first >> a.second;
  }
  template <typename A, typename B>
  ostream& operator<<(ostream& o, pair<A, B> a) {
    return o << '(' << a.first << ", " << a.second << ')';
  }
  namespace io {
  const long long SIZE = 1e7 + 10;
  char inbuff[SIZE];
  char *l, *r;
  inline void init() {
    l = inbuff;
    r = inbuff + fread(inbuff, 1, SIZE, stdin);
  }
  inline char gc() {
    if (l == r) init();
    return (l != r) ? *(l++) : EOF;
  }
  void R(long long& x) {
    x = 0;
    char ch = gc();
    while (!isdigit(ch)) ch = gc();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = gc();
  }
  }  // namespace io
  using io::R;
  void solve() {
    long long n, sum = 0;
    cin >> n;
    vector<long long> a(n);
    for (long long& i : a) cin >> i, sum += i;
    long long t = n * (n + 1) / 2;
    if (sum % t) {
      cout << "NO\n";
      return;
    }
    vector<long long> ans;
    sum /= t;
    for (long long i = 0; i <= n - 1; i++) {
      long long j = (i - 1 + n) % n;
      long long t = a[j] - a[i] + sum;
      if (t > 0 and t % n == 0)
        ans.push_back(t / n);
      else {
        cout << "NO\n";
        return;
      }
    }
    cout << "YES\n";
    for (auto i : ans) cout << i << ' ';
    cout << '\n';
    ;
  }
  void input() {}
  signed main() {
    ios_base::sync_with_stdio(false), cin.tie(0);
    ;
    long long tt = 1;
    cin >> tt;
    while (tt--) {
      input();
      solve();
    }
  }
  