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: //package codeforces.round760div3;
  
  import java.io.*;
  import java.util.*;
  
  import static java.lang.Math.*;
  
  //Think through the entire logic before jump into coding!
  //If you are out of ideas, take a guess! It is better than doing nothing!
  //Read both C and D, it is possible that D is easier than C for you!
  
  //Be aware of integer overflow!
  //If you find an answer and want to return immediately, don't forget to flush before return!
  
  public class E {
      static InputReader in;
      static PrintWriter out;
  
      public static void main(String[] args) {
          //initReaderPrinter(true);
          initReaderPrinter(false);
          solve(in.nextInt());
          //solve(1);        
      }
  
      static void solve(int testCnt) {
          for (int testNumber = 0; testNumber < testCnt; testNumber++) {
              int n = in.nextInt();
              int[] b = in.nextIntArrayPrimitive(n);
              long bSum = 0;
              for(int v : b) bSum += v;
              if(bSum % (1L * (n + 1) * n / 2) != 0) {
                  out.println("NO");
              }
              else {
                  long aSum = bSum / (1L * (n + 1) * n / 2);
                  long[] a = new long[n];
                  boolean can = true;
                  for(int i = 0; i < n; i++) {
                      long diff = aSum - (b[i] - b[(i - 1 + n) % n]);
                      if(diff <= 0 || diff % n != 0) {
                          can = false;
                          break;
                      }
                      a[i] = diff / n;
                  }
                  if(!can) out.println("NO");
                  else {
                      out.println("YES");
                      for(long v : a) {
                          out.print(v + " ");
                      }
                      out.println();
                  }
              }
          }
          out.close();
      }
  
      static void initReaderPrinter(boolean test) {
          if (test) {
              try {
                  in = new InputReader(new FileInputStream("src/input.in"));
                  out = new PrintWriter(new FileOutputStream("src/output.out"));
              } catch (IOException e) {
                  e.printStackTrace();
              }
          } else {
              in = new InputReader(System.in);
              out = new PrintWriter(System.out);
          }
      }
  
      static class InputReader {
          BufferedReader br;
          StringTokenizer st;
  
          InputReader(InputStream stream) {
              try {
                  br = new BufferedReader(new InputStreamReader(stream), 32768);
              } catch (Exception e) {
                  e.printStackTrace();
              }
          }
  
          String next() {
              while (st == null || !st.hasMoreTokens()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
  
          double nextDouble() {
              return Double.parseDouble(next());
          }
  
          String nextLine() {
              String str = "";
              try {
                  str = br.readLine();
              } catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
  
          Integer[] nextIntArray(int n) {
              Integer[] a = new Integer[n];
              for (int i = 0; i < n; i++) a[i] = nextInt();
              return a;
          }
  
          int[] nextIntArrayPrimitive(int n) {
              int[] a = new int[n];
              for (int i = 0; i < n; i++) a[i] = nextInt();
              return a;
          }
  
          int[] nextIntArrayPrimitiveOneIndexed(int n) {
              int[] a = new int[n + 1];
              for (int i = 1; i <= n; i++) a[i] = nextInt();
              return a;
          }
  
          Long[] nextLongArray(int n) {
              Long[] a = new Long[n];
              for (int i = 0; i < n; i++) a[i] = nextLong();
              return a;
          }
  
          long[] nextLongArrayPrimitive(int n) {
              long[] a = new long[n];
              for (int i = 0; i < n; i++) a[i] = nextLong();
              return a;
          }
  
          long[] nextLongArrayPrimitiveOneIndexed(int n) {
              long[] a = new long[n + 1];
              for (int i = 1; i <= n; i++) a[i] = nextLong();
              return a;
          }
  
          String[] nextStringArray(int n) {
              String[] g = new String[n];
              for (int i = 0; i < n; i++) g[i] = next();
              return g;
          }
  
          List<Integer>[] readGraphOneIndexed(int n, int m) {
              List<Integer>[] adj = new List[n + 1];
              for (int i = 0; i <= n; i++) {
                  adj[i] = new ArrayList<>();
              }
              for (int i = 0; i < m; i++) {
                  int u = nextInt();
                  int v = nextInt();
                  adj[u].add(v);
                  adj[v].add(u);
              }
              return adj;
          }
  
          List<Integer>[] readGraphZeroIndexed(int n, int m) {
              List<Integer>[] adj = new List[n];
              for (int i = 0; i < n; i++) {
                  adj[i] = new ArrayList<>();
              }
              for (int i = 0; i < m; i++) {
                  int u = nextInt() - 1;
                  int v = nextInt() - 1;
                  adj[u].add(v);
                  adj[v].add(u);
              }
              return adj;
          }
  
          /*
              A more efficient way of building a graph using int[] instead of ArrayList to store each node's neighboring nodes.
              1-indexed.
           */
          int[][] buildGraph(int nodeCnt, int edgeCnt) {
              int[] end1 = new int[edgeCnt], end2 = new int[edgeCnt];
              int[] edgeCntForEachNode = new int[nodeCnt + 1], idxForEachNode = new int[nodeCnt + 1];
              for (int i = 0; i < edgeCnt; i++) {
                  int u = in.nextInt(), v = in.nextInt();
                  edgeCntForEachNode[u]++;
                  edgeCntForEachNode[v]++;
                  end1[i] = u;
                  end2[i] = v;
              }
              int[][] adj = new int[nodeCnt + 1][];
              for (int i = 1; i <= nodeCnt; i++) {
                  adj[i] = new int[edgeCntForEachNode[i]];
              }
              for (int i = 0; i < edgeCnt; i++) {
                  adj[end1[i]][idxForEachNode[end1[i]]] = end2[i];
                  idxForEachNode[end1[i]]++;
                  adj[end2[i]][idxForEachNode[end2[i]]] = end1[i];
                  idxForEachNode[end2[i]]++;
              }
              return adj;
          }
      }
  }