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 z = 998244353;
  long long gcd(long long a, long long b) {
    if (a == 0) return b;
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long power(long long a, long long b) {
    long long res = 1;
    while (b) {
      if (b & 1) res = (res * a) % z, b--;
      a = (a * a) % z;
      b = b >> 1;
    }
    return res;
  }
  vector<long long> sv, pp;
  long long fact[200001];
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    pp.assign(1000001, 0);
    for (long long i = 2; i < 1000; i++) {
      if (!pp[i]) {
        for (long long j = i * i; j <= 1000000; j += i) {
          if (!pp[j]) pp[j] = 1;
        }
      }
    }
    fact[0] = 1;
    for (long long i = 1; i <= 200000; i++)
      fact[i] = i * fact[i - 1], fact[i] %= z;
    for (long long i = 2; i <= 1000000; i++)
      if (!pp[i]) sv.push_back(i);
    long long t;
    t = 1;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      long long ar[n], p = 0;
      for (long long i = 0; i < n; i++) {
        cin >> ar[i];
        p += ar[i];
      }
      long long k = (n * (n + 1)) / 2;
      if (p % k)
        cout << "NO\n";
      else {
        p /= k;
        long long pos = 1;
        vector<long long> ans(n, 0);
        for (long long i = 0; i < n - 1; i++) {
          long long p1 = -ar[i + 1] + ar[i];
          p1 += p;
          if (p1 % n) {
            pos = 0;
            break;
          } else {
            p1 /= n;
            ans[i + 1] = p1;
          }
        }
        long long p1 = ar[n - 1] - ar[0];
        p1 += p;
        if (p1 % n) {
          pos = 0;
        } else {
          p1 /= n;
          ans[0] = p1;
        }
        for (long long i = 0; i < n; i++) {
          if (ans[i] <= 0 || ans[i] > 1e9) {
            pos = 0;
            break;
          }
        }
        if (pos) {
          cout << "YES\n";
          for (long long i = 0; i < n; i++) cout << ans[i] << " ";
          cout << "\n";
        } else
          cout << "NO\n";
      }
    }
  }
  