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.*;
  import java.util.*;
  
  
  public class ComdeFormces {
  	public static void main(String[] args) throws Exception{
  		// TODO Auto-generated method stub
  		FastReader sc=new FastReader();
  		BufferedWriter log = new BufferedWriter(new OutputStreamWriter(System.out));
  		int t=sc.nextInt();
  		while(t--!=0) {
  		long n=sc.nextLong();
  		 long sum=(n*(n+1))/2;
  		 long sum2=0;
  		 ArrayList<Long> a=new ArrayList<>();
  		 for(int i=0;i<n;i++) {
  			 a.add(sc.nextLong());
  			 sum2+=a.get(i);
  		 }
  		 if(sum2%sum==0) {
  			 boolean ans=true;
  			 long val=sum2/sum;
  			 ArrayList<Long> arr=new ArrayList<>();
  			 long tsum=0;
  		 for(int i=0;i<n-1;i++) {
  			 long mnval=a.get(i)-a.get(i+1)+val;
  			 if(mnval%n!=0 || mnval<=0) {
  				 ans=false;
  				 break;
  			 }
  			 else {
  				 arr.add(mnval/n);
  				 tsum+=mnval/n;
  			 }
  			 
  		 }
  		if(val-tsum<=0)ans=false;
  		 if(ans) {
  			arr.add(0,val-tsum);
  			log.write("YES\n");
  			for(int i=0;i<arr.size();i++) {
  				log.write(arr.get(i)+" ");
  			}
  		 }
  		 else log.write("NO");
  		 
  		 }
  		 else log.write("NO");
           
  		
  		 
  		log.write("\n");
  		log.flush();
  		
  		}
  		
  	}
  static long find(int s,long a[]) {
  	if(s>=a.length)return -1;
  	long num=a[s];
  	for(int i=s;i<a.length;i+=2) {
  		num=gcd(num,a[i]);
  		if(num==1 || num==0)return -1;
  	}
  	return  num;
  }	
  static long gcd(long a,long b) {
  	if(b==0)return a;
  	else return gcd(b,a%b);
  }	
  static long factmod(long n,long mod,long img) {
  	if(n==0)return 1;
  	long ans=1;
  	long temp=1;
  	while(n--!=0) {
  		if(temp!=img) {
  		ans=((ans%mod)*((temp)%mod))%mod;
  		}
  		temp++;
  	}
  	return ans%mod;
  }	
  static long eval(ArrayList<Integer> a, int k) {
  	if(a.size()<1 )return 0;
  	int i=a.size()-1;
  	long ans=0;
  	while(i>=0) {
  		int temp=k;
  		int prev=0;
  		while(temp--!=0 && i>=0) {
  			ans+=Math.abs(a.get(i)-prev);
  			if(temp==0 || i==0)ans+=a.get(i);
  			prev=a.get(i);
  			i--;
  		}
  	}
  	return ans;
  }	
  static long bs(long end ,long cnt,long prev,long n) {
  	long start=1;
  	long st=end;
      while(start<=end) {
      	long mid=start+(end-start)/2;
      	if(mid*cnt+prev==n)return mid;
      	else if(mid*cnt+prev>n)end=mid-1;
      	else start=mid+1;
      }
      if(start<=st && start*cnt+prev>=n)return start;
  	return -1;
  }	
  	
  	
  //utils	
  public static class trip{
  	int a,b,c;
  	public trip(int a,int b,int c) {
  		this.a=a;
  		this.b=b;
  		this.c=c;
  	}
  	public int compareTo(trip q) {
  		return this.c-q.c;
  	}
  }	
  	static void mergesort(long[] a,int start,int end) {
  		if(start>=end)return;
  		int mid=start+(end-start)/2;
  		mergesort(a,start,mid);
  		mergesort(a,mid+1,end);
  		merge(a,start,mid,end);
  	}
  	static void merge(long []a, int start,int mid,int end) {
  		int ptr1=start;
  		int ptr2=mid+1;
  		long b[]=new long[end-start+1];
  		int i=0;
  		while(ptr1<=mid && ptr2<=end) {
  			if(a[ptr1]<=a[ptr2]) {
  				b[i]=a[ptr1];
  				ptr1++;
  				i++;
  			}
  			else {
  				b[i]=a[ptr2];
  				ptr2++;
  				i++;
  			}
  		}
  		while(ptr1<=mid) {
  			b[i]=a[ptr1];
  			ptr1++;
  			i++;
  		}
  		while(ptr2<=end) {
  			b[i]=a[ptr2];
  			ptr2++;
  			i++;
  		}
  		for(int j=start;j<=end;j++) {
  			a[j]=b[j-start];
  		}
  	}
  	public static class FastReader {
  
  		BufferedReader b;
  		StringTokenizer s;
  		public FastReader() {
  			b=new BufferedReader(new InputStreamReader(System.in));
  		}
  		String next() {
  			while(s==null ||!s.hasMoreElements()) {
  				try {
  					s=new StringTokenizer(b.readLine());
  				}
  				catch(IOException e) {
  					e.printStackTrace();
  				}
  			}
  			return s.nextToken();
  		}
  		public int nextInt() {
  			return Integer.parseInt(next());
  		}
  		public long nextLong() {
  			return Long.parseLong(next());
  		}
  		public double nextDouble() {
  			return Double.parseDouble(next());
  		}
  		String nextLine() {
  			String str="";
  			try {
  				str=b.readLine();
  			}
  			catch(IOException e) {
  				e.printStackTrace();
  			}
  			return str;
  		}
  		boolean hasNext() {
  		    if (s != null && s.hasMoreTokens()) {
  		        return true;
  		    }
  		    String tmp;
  		    try {
  		        b.mark(1000);
  		        tmp = b.readLine();
  		        if (tmp == null) {
  		            return false;
  		        }
  		        b.reset();
  		    } catch (IOException e) {
  		        return false;
  		    }
  		    return true;
  		}
  }
  public static class pair{
  	int a,b;
  	public pair(int a,int b) {
  		this.a=a;
  		this.b=b;
  	}
  	public int compareTo(pair b) {
  		return this.a-b.a;
  	}
  }	
  }
  