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: // Generated by Code Flattener.
  // https://plugins.jetbrains.com/plugin/9979-idea-code-flattener
  
  import java.io.Closeable;
  import java.io.DataInputStream;
  import java.io.IOException;
  
  public class E {
      public static void main(String[] args) throws IOException {
          FastScanner scanner = new FastScanner();
          int t = scanner.nextInt();
          cases:
          while (t-- > 0) {
              int n = scanner.nextInt();
  
              int[] b = new int[n];
              long sumB = 0;
              for (int i = 0; i < n; i++) {
                  b[i] = scanner.nextInt();
                  sumB += b[i];
              }
              if ((2 * sumB) % (n * (n + 1L)) != 0) {
                  System.out.println("NO");
                  continue cases;
              }
              long sum = 2 * sumB / n / (n + 1);
  
              int[] d = new int[n];
              d[0] = b[0] - b[n - 1];
              int max = Integer.MIN_VALUE;
              for (int i = 1; i < n; i++) {
                  d[i] = b[i] - b[i - 1];
                  max = Math.max(max, d[i]);
              }
              StringBuilder builder = new StringBuilder();
              for (int i = 0; i < n; i++) {
                  long val = sum - d[i];
                  if (val <= 0 || val % n != 0) {
                      System.out.println("NO");
                      continue cases;
                  }
                  builder.append(val / n);
                  if (i != n - 1) builder.append(" ");
              }
              System.out.println("YES");
              System.out.println(builder);
          }
      }
  
      private static class FastScanner implements Closeable {
          final private int BUFFER_SIZE = 1 << 16;
          private DataInputStream din;
          private byte[] buffer;
          private int bufferPointer, bytesRead;
  
          public FastScanner() {
              din = new DataInputStream(System.in);
              buffer = new byte[BUFFER_SIZE];
              bufferPointer = bytesRead = 0;
          }
  
          public int nextInt() throws IOException {
              int ret = 0;
              byte c = read();
              while (c <= ' ') c = read();
              boolean neg = (c == '-');
              if (neg) c = read();
              do {
                  ret = ret * 10 + c - '0';
              } while ((c = read()) >= '0' && c <= '9');
              if (neg) return -ret;
              return ret;
          }
  
          private void fillBuffer() throws IOException {
              bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
              if (bytesRead == -1) buffer[0] = -1;
          }
  
          private byte read() throws IOException {
              if (bufferPointer == bytesRead) fillBuffer();
              return buffer[bufferPointer++];
          }
  
          @Override
          public void close() throws IOException {
              if (din == null) return;
              din.close();
          }
      }
  }