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;
  bool dcmp(int x, int y) { return x > y; }
  bool acmp(int x, int y) { return x < y; }
  bool afcmp(pair<int, int> a, pair<int, int> b) { return a.first < b.first; }
  bool dfcmp(pair<int, int> a, pair<int, int> b) { return a.first > b.first; }
  bool ascmp(pair<int, int> a, pair<int, int> b) { return a.second < b.second; }
  bool dscmp(pair<int, int> a, pair<int, int> b) { return a.second > b.second; }
  int aabs(int a) { return a < 0 ? -a : a; }
  long long tmax(long long a, long long b, long long c) {
    return max(max(a, b), c);
  }
  long long tmin(long long a, long long b, long long c) {
    return min(min(a, b), c);
  }
  bool in_range(int x, int l, int r) { return x >= l && x <= r; }
  bool is_prime(int val) {
    if (val == 2) {
      return true;
    }
    if (val % 2 == 0) {
      return false;
    }
    for (int i = 3; i * i <= val; i += 2) {
      if (val % i == 0) {
        return false;
      }
    }
    return true;
  }
  int lowbit(int x) { return x & (-x); }
  inline long long gcd(long long x, long long y) { return y ? gcd(y, x % y) : x; }
  inline long long lcm(long long x, long long y) { return x / gcd(x, y) * y; }
  const int maxN = 4e4 + 5;
  bool finished = false;
  int N, a[maxN];
  long long ans[maxN];
  void reset() {
    cin >> N;
    for (int i = 1; i <= N; ++i) {
      cin >> a[i];
    }
  }
  void solve() {
    reset();
    if (finished) {
      return;
    }
    long long sum = 0, coeff = 0;
    for (int i = 1; i <= N; ++i) {
      sum += a[i];
      coeff += i;
    }
    if (sum % coeff != 0) {
      cout << "NO" << '\n';
      return;
    } else {
      sum /= coeff;
    }
    for (int i = 1; i <= N; ++i) {
      int diff = a[i] - a[i == 1 ? N : i - 1];
      long long x = sum - diff;
      if (x % N == 0 && x / N > 0) {
        ans[i] = x / N;
      } else {
        cout << "NO" << '\n';
        return;
      }
    }
    cout << "YES" << '\n';
    for (int i = 1; i <= N; ++i) {
      cout << ans[i] << " ";
    }
    cout << '\n';
  }
  int main(void) {
    std::ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int T = 1;
    cin >> T;
    while (T--) {
      solve();
    }
    return 0;
  }
  