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;
  template <typename A, typename B>
  ostream &operator<<(ostream &os, const pair<A, B> &p) {
    return os << '(' << p.first << ", " << p.second << ')';
  }
  template <typename T_container, typename T = typename enable_if<
                                      !is_same<T_container, string>::value,
                                      typename T_container::value_type>::type>
  ostream &operator<<(ostream &os, const T_container &v) {
    os << '{';
    string sep;
    for (const T &x : v) os << sep << x, sep = ", ";
    return os << '}';
  }
  void dbg_out() { cerr << endl; }
  template <typename Head, typename... Tail>
  void dbg_out(Head H, Tail... T) {
    cerr << ' ' << H;
    dbg_out(T...);
  }
  const long long MAX_N = 1e5 + 5;
  const long long MOD = 1e9 + 7;
  const long long INF = 1e9;
  const long double EPS = 1e-9;
  long long binpow(long long a, long long b) {
    if (b == 0) return 1;
    long long res = binpow(a, b / 2);
    if (b % 2)
      return res * res * a;
    else
      return res * res;
  }
  long long binpow(long long a, long long b, long long m) {
    a %= m;
    long long res = 1;
    while (b > 0) {
      if (b & 1) res = res * a % m;
      a = a * a % m;
      b >>= 1;
    }
    return res;
  }
  long long etf(long long n) {
    if (n == 1) return 1;
    long long phi = n;
    for (long long i = 2; i * i <= n; i++) {
      if (n % i == 0) {
        while (n % i == 0) n /= i;
        phi = phi - phi / i;
      }
    }
    if (n > 1) phi = phi - phi / n;
    return phi;
  }
  void solve() {
    long long n;
    cin >> n;
    long long b[n];
    long long f = (n * (n + 1)) / 2;
    long long sum = 0;
    for (long long i = 0; i < n; i++) {
      cin >> b[i];
      sum += b[i];
    }
    if (sum % f != 0) {
      cout << "NO\n";
      return;
    }
    long long suma = sum / f;
    long long a[n];
    a[0] = (suma - (b[0] - b[n - 1])) / n;
    if ((suma - (b[0] - b[n - 1])) % n != 0 || (suma - (b[0] - b[n - 1])) <= 0) {
      cout << "NO\n";
      return;
    }
    for (long long i = 1; i < n; i++) {
      long long dif = b[i] - b[i - 1];
      if ((suma - dif) % n != 0 || (suma - dif) <= 0) {
        cout << "NO\n";
        return;
      }
      a[i] = (suma - dif) / n;
    }
    cout << "YES\n";
    for (long long i = 0; i < n; i++) {
      cout << a[i] << ' ';
    }
    cout << '\n';
  }
  int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long tc = 1;
    cin >> tc;
    for (long long t = 1; t <= tc; t++) {
      solve();
    }
  }
  