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 M = 1e9 + 7;
  int biExp(long long int a, long long int n) {
    if (n == 0) return 1;
    long long int res = 1;
    while (n) {
      if (n & 1) res *= a, res %= M;
      a *= a, a %= M;
      n >>= 1;
    }
    return res;
  }
  int modMul(long long int a, long long int n) {
    if (n == 0) return 0;
    long long int res = 0;
    while (n) {
      if (n & 1) res += a, res %= M;
      a += a, a %= M;
      n >>= 1;
    }
    return res;
  }
  template <typename Arg1>
  void __f(const char *name, Arg1 &&arg1) {
    cout << name << " : " << arg1 << "\n";
  }
  template <typename Arg1, typename... Args>
  void __f(const char *names, Arg1 &&arg1, Args &&...args) {
    const char *comma = strchr(names + 1, ',');
    cout.write(names, comma - names) << " : " << arg1 << " || ";
    __f(comma + 1, args...);
  }
  void prakashitBhavah() {
    int n;
    cin >> n;
    long long int total_b = 0, total_a = 0;
    vector<int> b(n), a(n);
    for (int i = 0; i < n; i++) {
      cin >> b[i];
      total_b += b[i];
    }
    int x = n * (n + 1) / 2;
    if (total_b % x != 0) {
      cout << "NO\n";
      return;
    }
    total_a = total_b / x;
    bool flag = true;
    if (total_a < n) {
      cout << "NO\n";
      return;
    }
    long long int temp_a = 0;
    for (int i = 1; i < n; i++) {
      int diff = total_a - (b[i] - b[i - 1]);
      if (diff % n != 0) flag = false;
      a[i] = (diff) / n;
      temp_a += a[i];
    }
    a[0] = total_a - temp_a;
    for (int x : a) {
      if (x <= 0) flag = false;
    }
    if (flag) {
      cout << "YES\n";
      for (int x : a) cout << x << " ";
      cout << "\n";
    } else
      cout << "NO\n";
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--) prakashitBhavah();
    return 0;
  }
  