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 Solution {
  
  	static int mod = 1000000007;
  
  	public static void main(String[] args) {
  		FastScanner fs = new FastScanner();
  		int t = fs.nextInt();
  		outer: while (t-- > 0) {
  			int n = fs.nextInt();
  			int b[] = fs.readArray(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) {
  				System.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) {
  					System.out.println("NO");
  					continue outer;
  				}
  				a[next] = num / n;
  				if (a[next] <= 0) {
  					System.out.println("NO");
  					continue outer;
  				}
  			}
  			System.out.println("YES");
  			for (int i = 0; i < n; i++) {
  				System.out.print(a[i] + " ");
  			}
  			System.out.println();
  		}
  	}
  
  	static boolean isPrime(int n) {
  		if (n <= 1)
  			return false;
  		else if (n == 2)
  			return true;
  
  		else if (n % 2 == 0)
  			return false;
  
  		for (int i = 3; i <= Math.sqrt(n); i += 2) {
  			if (n % i == 0)
  				return false;
  		}
  		return true;
  	}
  
  	static int gcd(int a, int b) {
  		if (b == 0)
  			return a;
  		return gcd(b, a % b);
  	}
  
  	static long gcd(long a, long b) {
  		if (b == 0)
  			return a;
  		return gcd(b, a % b);
  	}
  
  	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());
  		}
  
  		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());
  		}
  
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  
  		ArrayList<Integer> readList(int n) {
  			ArrayList<Integer> list = new ArrayList<>();
  			for (int i = 0; i < n; i++)
  				list.add(nextInt());
  			return list;
  		}
  	}
  
  }
  