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.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.*;
  
  public class CF1 {
      public static void main(String[] args) {
          FastScanner sc=new FastScanner();
          int T=sc.nextInt();
  //        int T=1;
          for (int tt=0; tt<T; tt++){
              int n = sc.nextInt();
              long sum=0L;
              int arr[]= sc.readArray(n);
              for (int i=0; i<n; i++){
                  sum+=arr[i];
              }
              boolean possible =true;
              long y=(long)((long)n*(long)(n+1))/2;
              if (sum%y!=0) possible=false;
              sum/=y;
  
  //            if (sum<n)
  
              int ans[]= new int [n];
              for (int i=0; i<n-1; i++){
                  int x=arr[i]+(int)sum-arr[i+1];
                  if (x<=0 || x%n!=0) {
                      possible=false;
                      break;
                  }
                  else ans[i+1]=x/n;
              }
              int x =arr[n-1]+(int)sum-arr[0];
              if (x<=0 || x%n!=0) {
                  possible=false;
  
              }
              else ans[0]=x/n;
              if (possible){
                  System.out.println("YES");
                  StringBuilder sb= new StringBuilder();
                  for (int i=0; i<n; i++){
                      sb.append(ans[i]+" ");
                  }
                  System.out.println(sb.toString());
              }
              else System.out.println("NO");
          }
  // arr[i+1]= arr[0]+sum-n*x;
      }
      static boolean help(char arr[], int start, int end){
          for (int i=start+1; i<end; i++){
              if (arr[i]=='b' && arr[i-1]=='a' && arr[i+1]=='c') return true;
          }
          return false;
      }
      static long mod =998244353L;
      static long power (long a, long b){
          long res=1;
          while (b>0){
              if ((b&1)== 1){
                  res= (res * a % mod)%mod;
              }
              a=(a%mod * a%mod)%mod;
              b=b>>1;
          }
          return res;
      }
      boolean[] sieveOfEratosthenes(int n)
      {
  
          boolean prime[] = new boolean[n+1];
          for(int i=0;i<=n;i++)
              prime[i] = true;
  
          for(int p = 2; p*p <=n; p++)
          {
              if(prime[p] == true)
              {
                  for(int i = p*p; i <= n; i += p)
                      prime[i] = false;
              }
          }
          return prime;
      }
      static void sort(int[] a) {
          ArrayList<Integer> l=new ArrayList<>();
          for (int i:a) l.add(i);
          Collections.sort(l);
          for (int i=0; i<a.length; i++) a[i]=l.get(i);
      }
      static void sortLong(long[] a) {
          ArrayList<Long> l=new ArrayList<>();
          for (long i:a) l.add(i);
          Collections.sort(l);
          for (int i=0; i<a.length; i++) a[i]=l.get(i);
      }
      static long gcd (long n, long m){
          if (m==0) return n;
          else return gcd(m, n%m);
      }
      static class Pair{
          int x,y;
          public Pair(int x, int y){
              this.x = x;
              this.y = y;
          }
      }
      static class FastScanner {
          BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st=new StringTokenizer("");
          String next() {
              while (!st.hasMoreTokens())
                  try {
                      st=new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
          int[] readArray(int n) {
              int[] a=new int[n];
              for (int i=0; i<n; i++) a[i]=nextInt();
              return a;
          }
          long nextLong() {
              return Long.parseLong(next());
          }
      }
  }