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 MOD = 1000000007;
  long long int gcd(long long int a, long long int b) {
    return b ? gcd(b, a % b) : a;
  }
  unsigned long long power(unsigned long long x, long long int y,
                           long long int p) {
    unsigned long long res = 1;
    x = x % p;
    while (y > 0) {
      if (y & 1) res = (res * x) % p;
      y = y >> 1;
      x = (x * x) % p;
    }
    return res;
  }
  unsigned long long modInverse(unsigned long long n, long long int p) {
    return power(n, p - 2, p);
  }
  unsigned long long nCrModPFermat(unsigned long long n, long long int r,
                                   long long int p) {
    if (n < r) return 0;
    if (r == 0) return 1;
    unsigned long long fac[n + 1];
    fac[0] = 1;
    for (long long int i = 1; i <= n; i++) fac[i] = (fac[i - 1] * i) % p;
    return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
           p;
  }
  long long int find_set(long long int i, vector<long long int> &parent) {
    if (parent[i] == -1) {
      return i;
    } else {
      parent[i] = find_set(parent[i], parent);
      return parent[i];
    }
  }
  void union_set(long long int i, long long int j, vector<long long int> &parent,
                 vector<long long int> &rank) {
    long long int s1 = find_set(i, parent);
    long long int s2 = find_set(j, parent);
    if (s1 != s2) {
      if (rank[s1] > rank[s2]) {
        parent[s2] = s1;
        rank[s1] += rank[s2];
      } else {
        parent[s1] = s2;
        rank[s2] += rank[s1];
      }
    } else {
      return;
    }
  }
  void solve() {
    long long int n, i, j;
    cin >> n;
    long long int arr[n];
    long long int sum = 0;
    for (long long int i = 0; i < n; i++) {
      cin >> arr[i];
      sum += arr[i];
    }
    if ((2 * sum) % (n * (n + 1)) != 0) {
      cout << "NO" << endl;
      return;
    }
    long long int val = (2 * sum) / (n * (n + 1));
    if (val < n) {
      cout << "NO" << endl;
      return;
    }
    vector<long long int> a(n);
    for (i = 0; i < n; i++) {
      long long int val1 = arr[i];
      long long int val2 = arr[(i - 1 + n) % (n)];
      if ((val - val1 + val2) <= 0 || (val - val1 + val2) % n != 0) {
        cout << "NO" << endl;
        return;
      }
      a[i] = (val - val1 + val2) / n;
    }
    cout << "YES" << endl;
    for (i = 0; i < n; i++) {
      cout << a[i] << " ";
    }
    cout << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long int t;
    cin >> t;
    while (t--) solve();
  }
  