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;
  #pragma GCC optimize("O3")
  #pragma GCC target("avx2")
  template <typename T>
  void printv(const T& t) {
    std::copy(t.cbegin(), t.cend(),
              std::ostream_iterator<typename T::value_type>(std::cout, ", "));
    cout << '\n';
  }
  long long int modpower(long long int a, long long int b, long long int mod) {
    long long int res = 1;
    for (; b; b >>= 1) {
      if (b & 1) res = res * a;
      res %= mod;
      a = a * a;
      a %= mod;
    }
    return res;
  }
  struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
      x += 0x9e3779b97f4a7c15;
      x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
      x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
      return x ^ (x >> 31);
    }
    size_t operator()(uint64_t x) const {
      static const uint64_t FIXED_RANDOM =
          chrono::steady_clock::now().time_since_epoch().count();
      return splitmix64(x + FIXED_RANDOM);
    }
  };
  long long int expo(long long int a, long long int b, long long int mod) {
    long long int res = 1;
    while (b > 0) {
      if (b & 1) res = (res * a) % mod;
      a = (a * a) % mod;
      b = b >> 1;
    }
    return res;
  }
  long long int mod_inv(long long int a, long long int b) {
    return expo(a, b - 2, b);
  }
  long long int mod_add(long long int a, long long int b, long long int m) {
    a = a % m;
    b = b % m;
    return (((a + b) % m) + m) % m;
  }
  long long int mod_mul(long long int a, long long int b, long long int m) {
    a = a % m;
    b = b % m;
    return (((a * b) % m) + m) % m;
  }
  long long int mod_sub(long long int a, long long int b, long long int m) {
    a = a % m;
    b = b % m;
    return (((a - b) % m) + m) % m;
  }
  long long int mod_div(long long int a, long long int b, long long int m) {
    a = a % m;
    b = b % m;
    return (mod_mul(a, mod_inv(b, m), m) + m) % m;
  }
  void solve() {
    long long int n;
    cin >> n;
    vector<long long int> v(n);
    long long int sum = 0;
    for (long long int i = 0; i < n; i++) {
      cin >> v[i];
      sum += v[i];
    }
    long long int x = n * (n + 1);
    x /= 2;
    if (sum % x != 0) {
      cout << "NO\n";
      return;
    }
    long long int s = sum / x;
    vector<long long int> ans(n);
    for (long long int i = 1; i < n; i++) {
      long long int curr = s - (v[i] - v[i - 1]);
      if (curr <= 0 || curr % n != 0) {
        cout << "NO\n";
        return;
      }
      curr /= n;
      ans[i] = curr;
    }
    long long int curr = s - (v[0] - v[n - 1]);
    if (curr <= 0 || curr % n != 0) {
      cout << "NO\n";
      return;
    }
    curr /= n;
    ans[0] = curr;
    cout << "YES\n";
    for (long long int i = 0; i < n; i++) cout << ans[i] << " ";
    cout << '\n';
    return;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long int t;
    cin >> t;
    for (long long int i = 1; i <= t; i++) {
      solve();
    }
    return 0;
  }
  