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: import java.util.*;import java.io.*;import java.math.*;
  public class Main
  {
      public static void process(int t)throws IOException
      {
          int n = ni();
          long[]A = nal(n);
          long sum = 0;
          for(int i=0;i<n;i++) 
              sum += A[i];
          long den = (1l * n * (n + 1)) / 2;
          if(sum % den != 0){
              pn("NO");
              return;
          }
          sum /= den;
          long[]ans = new long[n];
          for(int i=1;i<n;i++){
              long term = sum + A[i-1]-A[i];
              if(term % n != 0 || term <= 0){
                  pn("NO");
                  return;
              }
              ans[i] = term/n;
          }
          ans[0] = A[0];
          for(int i=1;i<n;i++){
              ans[0] -= (n-i+1)*ans[i];
          }
          if(ans[0] <= 0){
              pn("NO");
              return;
          }
          pn("YES");
          for(long i:ans)p(i+" ");
          pn("");
      }
      static AnotherReader sc;
      static PrintWriter out;
      public static void main(String[]args)throws IOException
      {
          boolean oj = System.getProperty("ONLINE_JUDGE") != null;
          if(oj){sc=new AnotherReader();out=new PrintWriter(System.out);}
          else{sc=new AnotherReader(100);out=new PrintWriter("output.txt");}
          int t=1;
          t=ni();
          for(int i =1;i<=t;i++) {process(i);}
          out.flush();out.close();  
      }
  
      static void pn(Object o){out.println(o);}
      static void p(Object o){out.print(o);}
      static void pni(Object o){out.println(o);out.flush();}
      static int ni()throws IOException{return sc.nextInt();}
      static long nl()throws IOException{return sc.nextLong();}
      static double nd()throws IOException{return sc.nextDouble();}
      static String nln()throws IOException{return sc.nextLine();}
      static int[] nai(int N)throws IOException{int[]A=new int[N];for(int i=0;i!=N;i++){A[i]=ni();}return A;}
      static long[] nal(int N)throws IOException{long[]A=new long[N];for(int i=0;i!=N;i++){A[i]=nl();}return A;}
      static long gcd(long a, long b)throws IOException{return (b==0)?a:gcd(b,a%b);}
      static int gcd(int a, int b)throws IOException{return (b==0)?a:gcd(b,a%b);}
      static int bit(long n)throws IOException{return (n==0)?0:(1+bit(n&(n-1)));}
  
  /////////////////////////////////////////////////////////////////////////////////////////////////////////
  
      static class AnotherReader{BufferedReader br; StringTokenizer st;
      AnotherReader()throws FileNotFoundException{
      br=new BufferedReader(new InputStreamReader(System.in));}
      AnotherReader(int a)throws FileNotFoundException{
      br = new BufferedReader(new FileReader("input.txt"));}
      String next()throws IOException{
      while (st == null || !st.hasMoreElements()) {try{
      st = new StringTokenizer(br.readLine());}
      catch (IOException  e){ e.printStackTrace(); }}
      return st.nextToken(); } int nextInt() throws IOException{
      return Integer.parseInt(next());}
      long nextLong() throws IOException
      {return Long.parseLong(next());}
      double nextDouble()throws IOException { return Double.parseDouble(next()); }
      String nextLine() throws IOException{ String str = ""; try{
      str = br.readLine();} catch (IOException e){
      e.printStackTrace();} return str;}}
     
  /////////////////////////////////////////////////////////////////////////////////////////////////////////////
  }
  
  
  
  
  
  