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.InputStreamReader;
  import java.util.StringTokenizer;
  import java.util.*;
  import java.io.*;
  public class Main {
      // Graph
      // prefix sums
      //inputs
      public static void main(String args[])throws Exception{
          Input sc=new Input();
          precalculates p=new precalculates();
          StringBuilder sb=new StringBuilder();
          int t=sc.readInt();
          for(int f=0;f<t;f++){
              int n=sc.readInt();
              long a[]=sc.readArrayLong();
              long d[]=new long[n];
              long sum=0;
              for(int i=0;i<n;i++){
                  sum+=a[i];
              }
              long val=(long)n*(long)(n+1);
              val/=2;
              if(sum%val!=0){
                  sb.append("NO\n");
                  continue;
              }
              sum/=val;
              boolean b=false;
              long s2=0;
              for(int i=1;i<n;i++){
                  long temp=(sum-(a[i]-a[i-1]));
                  if(temp%n!=0 || temp<=0){
                      b=true;
                      break;
                  }else{
                      d[i]=temp/n;
                      s2+=d[i];
                  }
              }
              if(b || s2>=sum){
                  sb.append("NO\n");
                  continue;
              }
              sb.append("YES\n");
              d[0]=sum-s2;
              for(int i=0;i<n;i++){
                  sb.append(d[i]+" ");
              }
              sb.append("\n");
          }
          System.out.print(sb);
      }
      /*
      x 2x    x
      3y y    -2y
      2z 3z   z
      b2-b1=(x+y+z-(b2-b1))/3
       */
  }
  class Input{
      BufferedReader br;
      StringTokenizer st;
      Input(){
          br=new BufferedReader(new InputStreamReader(System.in));
          st=new StringTokenizer("");
      }
      public int[] readArray() throws Exception{
          st=new StringTokenizer(br.readLine());
          int a[]=new int[st.countTokens()];
          for(int i=0;i<a.length;i++){
              a[i]=Integer.parseInt(st.nextToken());
          }
          return a;
      }
      public long[] readArrayLong() throws Exception{
          st=new StringTokenizer(br.readLine());
          long a[]=new long[st.countTokens()];
          for(int i=0;i<a.length;i++){
              a[i]=Long.parseLong(st.nextToken());
          }
          return a;
      }
      public int readInt() throws Exception{
          st=new StringTokenizer(br.readLine());
          return Integer.parseInt(st.nextToken());
      }
      public long readLong() throws Exception{
          st=new StringTokenizer(br.readLine());
          return Long.parseLong(st.nextToken());
      }
      public String readString() throws Exception{
          return br.readLine();
      }
      public int[][] read2dArray(int n,int m)throws Exception{
          int a[][]=new int[n][m];
          for(int i=0;i<n;i++){
              st=new StringTokenizer(br.readLine());
              for(int j=0;j<m;j++){
                  a[i][j]=Integer.parseInt(st.nextToken());
              }
          }
          return a;
      }
  }
  class precalculates{
      public long gcd(long p, long q) {
          if (q == 0) return p;
          else return gcd(q, p % q);
      }
      public int[] prefixSumOneDimentional(int a[]){
          int n=a.length;
          int dp[]=new int[n];
          for(int i=0;i<n;i++){
              if(i==0)
                  dp[i]=a[i];
              else
                  dp[i]=dp[i-1]+a[i];
          }
          return dp;
      }
      public int[] postSumOneDimentional(int a[]) {
          int n = a.length;
          int dp[] = new int[n];
          for (int i = n - 1; i >= 0; i--) {
              if (i == n - 1)
                  dp[i] = a[i];
              else
                  dp[i] = dp[i + 1] + a[i];
          }
          return dp;
      }
      public int[][] prefixSum2d(int a[][]){
          int n=a.length;int m=a[0].length;
          int dp[][]=new int[n+1][m+1];
          for(int i=1;i<=n;i++){
              for(int j=1;j<=m;j++){
                  dp[i][j]=a[i-1][j-1]+dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1];
              }
          }
          return dp;
      }
      public long pow(long a,long b){
          long mod=1000000007;
          long ans=0;
          if(b<=0)
              return 1;
          if(b%2==0){
              ans=pow(a,b/2)%mod;
              return ((ans%mod)*(ans%mod))%mod;
          }else{
              return ((a%mod)*(ans%mod))%mod;
          }
      }
  
  }
  class GraphInteger{
      HashMap<Integer,vertex> vtces;
      class vertex{
          HashMap<Integer,Integer> children;
          public vertex(){
              children=new HashMap<>();
          }
      }
      public GraphInteger(){
          vtces=new HashMap<>();
      }
      public void addVertex(int a){
          vtces.put(a,new vertex());
      }
      public void addEdge(int a,int b,int cost){
          if(!vtces.containsKey(a)){
              vtces.put(a,new vertex());
          }
          if(!vtces.containsKey(b)){
              vtces.put(b,new vertex());
          }
          vtces.get(a).children.put(b,cost);
  //        vtces.get(b).children.put(a,cost);
      }
      public boolean isCyclicDirected(){
          boolean isdone[]=new boolean[vtces.size()+1];
          boolean check[]=new boolean[vtces.size()+1];
          for(int i=1;i<=vtces.size();i++) {
              if (!isdone[i] && isCyclicDirected(i,isdone, check)) {
                  return true;
              }
          }
          return false;
      }
      private boolean isCyclicDirected(int i,boolean isdone[],boolean check[]){
          if(check[i])
              return true;
          if(isdone[i])
              return false;
          check[i]=true;
          isdone[i]=true;
          Set<Integer> set=vtces.get(i).children.keySet();
          for(Integer ii:set){
              if(isCyclicDirected(ii,isdone,check))
                  return true;
          }
          check[i]=false;
          return false;
      }
  }
  class union_find {
      int n;
      int[] sz;
      int[] par;
  
      union_find(int nval) {
          n = nval;
          sz = new int[n + 1];
          par = new int[n + 1];
          for (int i = 0; i <= n; i++) {
              par[i] = i;
              sz[i] = 1;
          }
      }
  
      int root(int x) {
          if (par[x] == x)
              return x;
          return par[x] = root(par[x]);
      }
  
      boolean find(int a, int b) {
          return root(a) == root(b);
      }
  
      int union(int a, int b) {
          int ra = root(a);
          int rb = root(b);
          if (ra == rb)
              return 0;
          if(a==b)
              return 0;
          if (sz[a] > sz[b]) {
              int temp = ra;
              ra = rb;
              rb = temp;
          }
          par[ra] = rb;
          sz[rb] += sz[ra];
          return 1;
      }
  }
  
  
  
  
  
  
  