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>
  #pragma optimize(3, "-Ofast", "inline")
  using namespace std;
  const int maxn = 1e5 + 3, maxm = 1e6 + 11, inf_int = 0x3f3f3f3f;
  const long long inf_ll = 0x3f3f3f3f3f3f3f, mod = 1e9 + 7;
  const double eps = 1e-6;
  long long n, a[maxn], b[maxn], d[maxn];
  inline long long get(long long x0) {
    long long x1 = x0 - 1, x2 = x0 - 2;
    if (x1 <= 0) x1 += n;
    if (x2 <= 0) x2 += n;
    return 2 * b[x1] - b[x0] - b[x2];
  }
  inline void solve() {
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> b[i];
    for (int i = 1; i <= n; i++) {
      d[i] = get(i);
      if (d[i] % n) {
        cout << "NO" << endl;
        return;
      }
      d[i] /= n;
    }
    a[1] = 0;
    for (int i = 2; i <= n; i++) a[i] = d[i] + a[i - 1];
    long long sum = b[1], tot = n * (n + 1) / 2;
    for (int i = 2, t = n; i <= n; i++, t--) sum -= t * a[i];
    if (sum % tot) {
      cout << "NO" << endl;
      return;
    }
    a[1] = sum / tot;
    for (int i = 2; i <= n; i++) a[i] += a[1];
    for (int i = 1; i <= n; i++) {
      if (a[i] <= 0) {
        cout << "NO" << endl;
        return;
      }
    }
    cout << "YES" << endl;
    for (int i = 1; i <= n; i++) cout << a[i] << " ";
    cout << endl;
  }
  int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int TEST = 1;
    cin >> TEST;
    while (TEST--) solve();
  }
  