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.util.*;
  import java.io.*;
  
  public class path {
  
  	static boolean check(long d, long[] arr) {
  		if (d == 0)
  			return true;
  		for (int i = 1; i < arr.length; i++) {
  			if (arr[i - 1] % d == 0 && arr[i] % d == 0)
  				return false;
  			if (arr[i - 1] % d != 0 && arr[i] % d != 0)
  				return false;
  		}
  		return true;
  	}
  
  	static long gcd(long a, long b) {
  		if (b == 0)
  			return a;
  		else
  			return gcd(b, a % b);
  	}
  
  	/*
  	 * static long binarySearch(long start, long end, long [] arr){ if(arr.length ==
  	 * 2) return Math.max(arr[0], arr[1]);
  	 * 
  	 * while(start < end){ long mid = (start + end) / 2; if(check(mid, arr)){ ans =
  	 * mid; start = mid + 1; } else end = mid; }
  	 * 
  	 * for(int i = 0; i < arr.length - 2; i += 2) { long p = gcd(arr[i], arr[i +
  	 * 2]); if(p != 1) { if(check(p, arr)) return p; } } for(int i = 1; i <
  	 * arr.length - 2; i += 2) { long p = gcd(arr[i], arr[i + 2]); if(p != 1) {
  	 * if(check(p, arr)) return p; } } return 0; }
  	 */
  	
  	public static void main(String[] args) throws IOException {
  		BufferedReader sc = new BufferedReader(new InputStreamReader(System.in));
  		int q = Integer.parseInt(sc.readLine());
  		for (int t = 0; t < q; t++) {
  			int n = Integer.parseInt(sc.readLine());
  			String [] s = sc.readLine().split(" ");
  			int [] arr = new int [n];
  			long sum = 0;
  			for(int i = 0; i < n; i++) {
  				arr[i] = Integer.parseInt(s[i]);
  				sum += arr[i];
  			}
  			long pre = (n * (n + 1)) / 2;
  			boolean flag = false;
  			if(sum % pre == 0 && sum / pre >= n) {
  				
  				String toPrint = "";
  				for(int i = 0; i < n; i++) {
  					if(i == n - 1) {
  						long sub1 = arr[n - 1] - arr[0] + (sum / pre);
  						if(sub1 % n != 0 || sub1 <= 0) {
  							flag = true;
  							break;
  						}
  						else {
  							toPrint = Long.toString(sub1 / n) + " " + toPrint;
  						}
  					}
  					else {
  						long sub1 = arr[i] - arr[i + 1] + (sum / pre);
  						if(sub1 % n != 0 || sub1 <= 0) {
  							flag = true;
  							break;
  						}
  						else {
  							toPrint = toPrint + Long.toString(sub1 / n) + " ";
  						}
  					}
  				}
  				if(flag) {
  					System.out.println("NO");
  					continue;
  				}
  				System.out.println("YES");
  				System.out.println(toPrint);
  			}
  			else {
  				System.out.println("NO");
  			}
  		}
  	}
  }
  