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;
  long long int fastpow(long long int a, long long int p) {
    long long int res = 1;
    while (p) {
      if (p % 2 == 0) {
        a = a * 1ll * a % 1000000007;
        p /= 2;
      } else {
        res = res * 1ll * a % 1000000007;
        p--;
      }
    }
    return res;
  }
  long long int fact(long long int n) {
    long long int res = 1;
    for (long long int i = 1; i <= n; i++) {
      res = res * 1ll * i % 1000000007;
    }
    return res;
  }
  long long int comb(long long int n, long long int k) {
    return fact(n) * 1ll * fastpow(fact(k), 1000000007 - 2) % 1000000007 * 1ll *
           fastpow(fact(n - k), 1000000007 - 2) % 1000000007;
  }
  long long int to_integer(string& aa) {
    long long int res = 0;
    for (auto x : aa) res = (res * 10) + (x - '0');
    return res;
  }
  bool prime(long long int n) {
    if (n <= 1) return false;
    if (n <= 3) return true;
    if (n % 2 == 0 || n % 3 == 0) return false;
    for (long long int i = 5; i * i <= n; i = i + 6)
      if (n % i == 0 || n % (i + 2) == 0) return false;
    return true;
  }
  void sieve(long long int n, vector<long long int>& prime) {
    for (long long int p = 2; p * p <= n; p++) {
      if (prime[p] == 1) {
        for (long long int i = p * p; i <= n; i += p) prime[i] = 0;
      }
    }
  }
  bool check(string s, string a) {
    long long int j = 0;
    for (long long int i = 0; i < s.size(); i++) {
      if (s[i] == a[j]) {
        j++;
      }
      if (j == a.size()) return true;
    }
    return false;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long int t;
    cin >> t;
    while (t--) {
      long long int n;
      cin >> n;
      vector<long long int> b(n);
      long long int s = 0;
      for (long long int i = 0; i < n; i++) {
        cin >> b[i];
        s += b[i];
      }
      long long int x = n * (n + 1) / 2;
      if (s % x != 0) {
        cout << "NO\n";
        continue;
      }
      long long int f = 0;
      s /= x;
      vector<long long int> ans(n);
      for (long long int i = 0; i < n; i++) {
        if (i != n - 1) {
          long long int tmp = b[i] - b[i + 1] + s;
          if (tmp <= 0 || tmp % n != 0) {
            f = 1;
            break;
          }
          ans[i + 1] = tmp / n;
        } else {
          long long int tmp = b[i] - b[0] + s;
          if (tmp <= 0 || tmp % n != 0) {
            f = 1;
            break;
          }
          ans[0] = tmp / n;
        }
      }
      if (f) {
        cout << "NO\n";
      } else {
        cout << "YES\n";
        for (auto x : ans) {
          cout << x << " ";
        }
        cout << "\n";
      }
    }
  }
  