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: #from itertools import *
  #from math import *
  #from bisect import *
  #from collections import *
  #from random import *
  #from decimal import *
  #from heapq import *
  #from itertools import *            # Things Change ....remember :)
  import sys
  input=sys.stdin.readline
  def inp():
      return int(input())
  def st():
      return input().rstrip('\n')
  def lis():
      return list(map(int,input().split()))
  def ma():
      return map(int,input().split())
  def painalways(a,b):
      if(len(b)<len(a)):
          return 0
      for i in range(len(b)-len(a)+1):
          if(b[i:i+len(a)]==a and b[:i].count('0')==0 and b[i+len(a):].count('0')==0):
              return 1
  t=inp()
  zz=10**9
  while(t):
      t-=1
      n=inp()
      a=lis()
      if(n==1):
          print("YES")
          print(*a)
      else:
          flag=0
          x=n*(n+1)
          x=x//2
          if(sum(a)%x):
              print("NO")
          else:
              final=sum(a)//x
              check=[0]*n
              for i in range(n-1):
                  diff=a[i+1]-a[i]
                  tem=final-diff
                  if(tem%n or tem==0):
                      flag=1
                      break
                  check[i+1]=tem//n
              diff=a[0]-a[-1]
              tem=final-diff
              if(tem%n or tem==0):
                  flag=1
              check[0]=tem//n
              for i in check:
                  if(i>zz or i<=0):
                      flag=1
              if(flag):
                  print("NO")
              else:
                  print("YES")
                  print(*check)
                  
  