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: /*
                         _oo0oo_
                        o8888888o
                        88" . "88
                        (| -_- |)
                        0\  =  /0
                      ___/`---'\___
                    .' \\|     |// '.
                   / \\|||  :  |||// \
                  / _||||| -:- |||||- \
                 |   | \\\  -  /// |   |
                 | \_|  ''\---/''  |_/ |
                 \  .-\__  '-'  ___/-. /
               ___'. .'  /--.--\  `. .'___
            ."" '<  `.___\_<|>_/___.' >' "".
           | | :  `- \`.;`\ _ /`;.`/ - ` : | |
           \  \ `_.   \_ __\ /__ _/   .-` /  /
       =====`-.____`.___ \_____/___.-`___.-'=====
                         `=---='
   
  */
  
  import java.util.*;
  import java.math.*;
  import java.io.*;
  import java.lang.Math.*;
  
  public class KickStart2020 {
  	 static class FastReader {
  		BufferedReader br;
  		StringTokenizer st;
  
  		public FastReader() {
  			br = new BufferedReader(new InputStreamReader(System.in));
  		}
  
  		String next() {
  			while (st == null || !st.hasMoreElements()) {
  				try {
  					st = new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			}
  			return st.nextToken();
  		}
  
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  
  		float nextFloat() {
  			return Float.parseFloat(next());
  		}
  
  		String nextLine() {
  			String str = "";
  			try {
  				str = br.readLine();
  			} catch (IOException e) {
  				e.printStackTrace();
  			}
  			return str;
  		}
  	}
  
  	static long gcd(long a, long b) {
  		if (b == 0)
  			return a;
  		return gcd(b, a % b);
  
  	}
  	static long lcm(long a, long b) {
  		return a / gcd(a, b) * b;
  	}
  
  	public static class Pair implements Comparable<Pair> {
  		public int index;
  		public String value;
  
  		public Pair(int index,String value) {
  			this.index = index;
  			this.value = value;
  		}
  
  		@Override
  		public int compareTo(Pair other) {
  			// multiplied to -1 as the author need descending sort order
  			if(other.index < this.index) return -1;
  			if(other.index > this.index) return 1;
  			return 0;
  		}
  		@Override
  		public String toString() {
  			return this.index + " " + this.value;
  		}
  	}
  	static boolean isPrime(long d) {
  		if (d == 3)
  			return false;
  		for (int i = 2; i <= (long) Math.sqrt(d); i++) {
  			if (d % i == 0)
  				return false;
  		}
  		return true;
  	}
  	static void decimalTob(int n, int k, Stack<Integer> ss) {
            int x = n % k;
            int y = n / k;
            ss.push(x);
            if(y > 0) {
          	   decimalTob(y, k, ss);
            }
  	}
  	static long powermod(long x, long y, long mod) {
  		long ans = 1;
  		x = x % mod;
  		if (x == 0)
  			return 0;
  		int i = 1;
  		while (y > 0) {
  			if ((y & 1) != 0)
  				ans = (ans * x) % mod;
  			y = y >> 1;
  			x = (x * x) % mod;
  		}
  		return ans;
  	}
  	static long power(long x, long y)
  	{
  	    long res = 1;
  	    while (y > 0)
  	    {
  	        if ((y & 1) != 0)
  	            res = res * x;	  
  	        y = y >> 1; 
  	        x = x * x;  
  	    }
  	    return res;
  	}
  	public static void main(String[] args) throws Exception {
  		FastReader sc = new FastReader();
  		PrintWriter out = new PrintWriter(System.out);
  		int t = sc.nextInt();
  		outerloop:
  			while(t-- > 0) {
               int n = sc.nextInt();
               int brr[] = new int[n];
               long sum = 0;
               for(int i = 0; i < n; i++) {
              	 sum += brr[i] = sc.nextInt();
               }
               int total = n * (n + 1) / 2;
               if(sum % total == 0 && sum / total >= n) {
              	 long z = sum / total;
              	 long arr[] = new long[n];
              	 for(int i = 0; i < n; i++) {
              		 long k = brr[i] - brr[(i + 1) % n] + z;
              		 if(k > 0 && k % n == 0) arr[(i + 1) % n] = k / n;
              		 else {
              			 out.println("NO");
              			 continue outerloop;
              		 }
              	 }
              	 out.println("YES");
              	 for(int i = 0; i < n; i++) out.print(arr[i] + " ");
              	 out.println();
               }
               else out.println("NO");
  			}
  		out.close();
  	}
  }