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;
  void ha() { cout << "YES" << endl; }
  void na() { cout << "NO" << endl; }
  long long power(long long a, long long b) {
    long long c, d;
    if (b == 0) {
      return 1 % 1000000007;
    } else if (b == 1) {
      return a % 1000000007;
    } else if (b % 2 == 0) {
      c = power(a, b / 2);
      return (c * c) % 1000000007;
    } else {
      c = power(a, b / 2);
      d = (c * c) % 1000000007;
      d = (d * a) % 1000000007;
      return d;
    }
  }
  long long plainPower(long long a, long long b) {
    long long sum = 1, i;
    for (i = 1; i <= b; i++) {
      sum *= a;
    }
    return sum;
  }
  int absolute(int a) {
    if (a >= 0) {
      return a;
    } else {
      return (a * (-1));
    }
  }
  long long absolute(long long a) {
    if (a >= 0) {
      return a;
    } else {
      return (a * (-1));
    }
  }
  bool isPrime(int a) {
    for (int i = 2; i * i <= a; i++) {
      if (a % i == 0) {
        return false;
      }
    }
    return true;
  }
  bool isTriangle(long long x1, long long y1, long long x2, long long y2,
                  long long x3, long long y3) {
    long long area = (x1 * (y2 - y3)) + (x2 * (y3 - y1)) + (x3 * (y1 - y2));
    if (area) {
      return true;
    } else {
      return false;
    }
  }
  long long findRangeSum(long long hi, long long lo) {
    long long j = (hi * (hi + 1)) / 2;
    long long k = (lo * (lo + 1)) / 2;
    return j - k;
  }
  bool present(vector<int> &a, int n, int target) {
    int lo = 0, hi = n - 1, mid;
    while (lo <= hi) {
      mid = lo + ((hi - lo) / 2);
      if (a[mid] == target) {
        return true;
      } else if (a[mid] > target) {
        hi = mid - 1;
      } else {
        lo = mid + 1;
      }
    }
    return false;
  }
  int Left_bound(vector<int> &a, int n, int target) {
    int lo = 0, hi = n - 1, mid, ans = -1;
    while (lo <= hi) {
      mid = lo + ((hi - lo) / 2);
      if (a[mid] <= target) {
        ans = mid;
        lo = mid + 1;
      } else {
        hi = mid - 1;
      }
    }
    return ans + 1;
  }
  int Right_bound(vector<int> &a, int n, int target) {
    int lo = 0, hi = n - 1, mid, ans = n;
    while (lo <= hi) {
      mid = lo + ((hi - lo) / 2);
      if (a[mid] >= target) {
        ans = mid;
        hi = mid - 1;
      } else {
        lo = mid + 1;
      }
    }
    return ans + 1;
  }
  void debugger() {}
  int cnt;
  int target;
  int dfs(vector<vector<int> > &adj, vector<int> &vis, vector<int> &val, int strt,
          int parent) {
    vis[strt] = val[strt];
    int xr = 0;
    for (int i = 0; i < adj[strt].size(); i++) {
      if (adj[strt][i] != parent) {
        xr = (xr ^ dfs(adj, vis, val, adj[strt][i], strt));
      }
    }
    vis[strt] = (xr ^ vis[strt]);
    if (vis[strt] == target) {
      cnt++;
      vis[strt] = 0;
    }
    return vis[strt];
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      long long term = (n * (n + 1)) / 2;
      vector<long long> a(n + 1);
      vector<long long> temp(n + 1);
      long long i, j, k, flag = 0, sum = 0;
      for (i = 0; i < n; i++) {
        cin >> a[i];
        sum += a[i];
        temp[i] = 0;
      }
      if (sum % term) {
        na();
        continue;
      }
      long long tt = sum / term;
      vector<long long> arr;
      for (i = 1; i < n; i++) {
        arr.push_back(a[i] - a[i - 1]);
      }
      arr.push_back(a[0] - a[n - 1]);
      for (i = 1; i <= n; i++) {
        j = i % n;
        temp[j] = tt - arr[i - 1];
        if (temp[j] % n) {
          flag = 1;
          break;
        }
        temp[j] /= n;
        if (temp[j] <= 0) {
          flag = 1;
          break;
        }
      }
      if (flag) {
        na();
      } else {
        ha();
        for (i = 0; i < n; i++) {
          cout << temp[i] << " ";
        }
        cout << endl;
      }
    }
    return 0;
  }
  