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 N = 2e5 + 7;
  long long A[N], sum[N], dif[N], dif1[N], ans[N];
  void solve() {
    int n;
    cin >> n;
    for (int i = 1; i <= n; ++i) {
      cin >> A[i];
      if (i > 1) sum[i] = A[i] - A[i - 1];
    }
    sum[1] = A[1] - A[n];
    for (int i = 2; i <= n; ++i) {
      dif[i] = sum[i - 1] - sum[i];
      if (dif[i] % n) {
        cout << "NO\n";
        return;
      }
      dif[i] /= n;
    }
    for (int i = 2; i <= n; ++i) {
      dif1[i] = dif1[i - 1] + dif[i];
    }
    long long k = 1, b = 0;
    for (int i = 2; i <= n; ++i) {
      k += (n - i + 2);
      b += dif1[i] * (n - i + 2);
    }
    if ((A[1] - b) % k) {
      cout << "NO\n";
      return;
    }
    ans[1] = (A[1] - b) / k;
    for (int i = 1; i <= n; ++i) {
      ans[i] = ans[1] + dif1[i];
      if (ans[i] < 1) {
        cout << "NO\n";
        return;
      }
    }
    cout << "YES\n";
    for (int i = 1; i <= n; ++i) cout << ans[1] + dif1[i] << ' ';
    cout << '\n';
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int t = 1;
    cin >> t;
    while (t--) solve();
  }
  