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;
  long long N = 1e9 + 7;
  void solve() {
    long long n;
    cin >> n;
    vector<long long> v(n);
    for (int i = 0; i < n; i++) cin >> v[i];
    long long b[n];
    long long sum = 0;
    for (int i = 0; i < n; i++) sum += v[i];
    long long x = (n * (n + 1)) / 2;
    long long tot = sum / x;
    if (sum % x != 0) {
      cout << "NO";
      return;
    }
    b[0] = v[n - 1] - v[0];
    for (int i = 1; i < n; i++) {
      b[i] = v[i - 1] - v[i];
    }
    long long ans = 0;
    vector<long long> an;
    for (int i = 0; i < n; i++) {
      ans = (b[i] + tot) / (n);
      an.push_back(ans);
    }
    if ((b[0] + tot) % n != 0) an[0] = -1;
    int f = 1;
    for (int i = 0; i < n; i++) {
      if (an[i] <= 0) f = 0;
    }
    if (f) {
      cout << "YES\n";
      for (auto y : an) cout << y << " ";
    } else {
      cout << "NO";
    }
  }
  int main() {
    int t;
    cin >> t;
    while (t--) {
      solve();
      cout << '\n';
    }
    return 0;
  }
  