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.io.PrintWriter;
  import java.text.DecimalFormat;
  import java.util.*;
  
  
   
  public class Codeforces {
  	
  	static int mod= 1000000007;
  	
  	public static void main(String[] args) throws Exception {
  		PrintWriter out=new PrintWriter(System.out);
  		FastScanner fs=new FastScanner();
  		int t=fs.nextInt();
  		outer:while(t-->0) {
  			int n=fs.nextInt();
  			long b[]=fs.lreadArray(n);
  			long sum=0;
  			for(int i=0;i<n;i++) sum+=b[i];
  			long a[]=new long[n];
  			long nn= (n*(long)(n+1))/2;
  			if(sum%nn!=0) {
  				out.println("NO");
  				continue outer;
  			}
  			long sa= sum/nn;
  			for(int i=0;i<n;i++) {
  				int cur=i, next= (i+1)%n;
  				long num= b[cur]-b[next]+sa;
  				if(num%n!=0) {
  					out.println("NO");
  					continue outer;
  				}
  				a[next]= num/n;
  				if(a[next]<=0) {
  					out.println("NO");
  					continue outer;
  				}
  			}
  			out.println("YES");
  			for(int i=0;i<n;i++) {
  				out.print(a[i]+" ");
  			}
  			out.println();
  		}
  		out.close();
  		
  	}
  	
  	static long pow(long a,long b) {
  		if(b<=0) return 1;
  		long res=1;
  		while(b!=0) {
  			if((b&1)!=0) {
  				res*=a;
  				res%=mod;
  			}
  			a*=a;
  			a%=mod;
  			b=b>>1;
  		}
  		return res;
  	}
  	static long gcd(long  a,long  b) {
  		if(b==0) return a;
  		return gcd(b,a%b);
  	}
  	static long nck(int n,int k) {
  		if(k>n) return 0;
  		long res=1;
  		res*=fact(n);
  		res%=mod;
  		res*=modInv(fact(k));
  		res%=mod;
  		res*=modInv(fact(n-k)); 
  		res%=mod;
  		return res;
  	}
  	static long fact(long n) {
  //		return fact[(int)n];
  		long res=1;
  		for(int i=2;i<=n;i++) {
  			res*=i;
  			res%=mod;
  		}
  		return res;
  	}
  	
  	static long modInv(long n) {
  		return pow(n,mod-2);
  	}
  	
  	static void sort(int[] a) {
  		//suffle
  		int n=a.length;
  		Random r=new Random();
  		for (int i=0; i<a.length; i++) {
  			int oi=r.nextInt(n);
  			int temp=a[i];
  			a[i]=a[oi];
  			a[oi]=temp;
  		}
  		
  		//then sort
  		Arrays.sort(a);
  	}
  	
  	// Use this to input code since it is faster than a Scanner
  	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());
  		}
  		long[] lreadArray(int n) {
  			long a[]=new long[n];
  			for(int i=0;i<n;i++) a[i]=nextLong();
  			return a;
  		}
  		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());
  		}
  	}
   
  }