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;
  using ll = long long;
  using ld = long double;
  using str = string;
  using pi = pair<int, int>;
  using pl = pair<ll, ll>;
  using pd = pair<ld, ld>;
  template <class T>
  using V = vector<T>;
  using vi = V<int>;
  using vb = V<bool>;
  using vl = V<ll>;
  using vd = V<ld>;
  using vs = V<str>;
  using vpi = V<pi>;
  using vpl = V<pl>;
  using vpd = V<pd>;
  template <class T>
  int lwb(V<T>& a, const T& b) {
    return int(lower_bound(begin(a), end(a), b) - bg(a));
  }
  template <class T>
  int upb(V<T>& a, const T& b) {
    return int(upper_bound(begin(a), end(a), b) - bg(a));
  }
  const int MOD = 1e9 + 7;
  const int MX = 1e5 + 5;
  const ll BIG = 1e18;
  const int dx[4]{1, 0, -1, 0}, dy[4]{0, 1, 0, -1};
  mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());
  template <class T>
  using pqg = priority_queue<T, vector<T>, greater<T>>;
  constexpr int pct(int x) { return __builtin_popcount(x); }
  constexpr int bits(int x) { return x == 0 ? 0 : 31 - __builtin_clz(x); }
  constexpr int p2(int x) { return 1 << x; }
  ll cdiv(ll a, ll b) { return a / b + ((a ^ b) > 0 && a % b); }
  ll fdiv(ll a, ll b) { return a / b - ((a ^ b) < 0 && a % b); }
  template <class T>
  bool ckmin(T& a, const T& b) {
    return b < a ? a = b, 1 : 0;
  }
  template <class T>
  bool ckmax(T& a, const T& b) {
    return a < b ? a = b, 1 : 0;
  }
  template <class T>
  ll gcd(T& a, T& b) {
    if (a > b) swap(a, b);
    while (a) {
      b -= a;
      if (a > b) swap(a, b);
    }
    return b;
  }
  template <class T>
  ll power(T& a, T& b) {
    ll ans = 1;
    while (b) {
      if (b & 1) ans *= a;
      a *= a;
      b >>= 1;
    }
    return ans;
  }
  void solve() {
    int n;
    cin >> n;
    int b[n], a[n];
    ll sum = 0;
    for (int i = (0); i < (n); ++i) {
      cin >> b[i];
      sum += b[i];
    }
    ll cnt = n * 1ll * (n + 1) / 2;
    if (sum % cnt != 0) {
      cout << "NO\n";
      return;
    }
    for (int i = (n)-1; i >= (0); --i) {
      ll res = b[i] - b[(i + 1) % n] + (sum / cnt);
      if (res % n != 0 || res <= 0) {
        cout << "NO\n";
        return;
      }
      a[(i + 1) % n] = res / n;
    }
    cout << "YES\n";
    for (int i = (0); i < (n); ++i) cout << a[i] << " ";
    cout << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--) solve();
  }
  