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 sys
  
  def readstr():
      input = sys.stdin.readline
      a=str(input().split()[0])
      return a
  
  def readnum():
      input = sys.stdin.readline
      a=list(map(int,input().split()))
      return a
  
  def printf(val):
      sys.stdout.write(str(val) + '\n')
  
  def power(a, b, mod):
      if b == 1:
          return a
      if b == 0:
          return 1
  
      ans = power(a, b//2, mod)
      ans *= ans
  
      if b % 2:
          ans *= a
      
      ans = ans % mod
      return ans
  
  
  def factorial(n, start=1, skip=None):
      res = start
      for i in range(1, n+1):
          if i == skip:
              continue
          res *= i
  #        res = res % mod
  
      return res
  
  def make_deg(l):
      deg = {}
      for i in l:
          if not i in deg:
              deg[i] = 0
          deg[i] += 1
      return deg
  
  for _ in range(readnum()[0]):
      n = readnum()[0]
      b = readnum()
  
      if (2*sum(b)) % (n*(n+1)):
          print ("NO")
          continue
  
      s = 2*sum(b)//n//(n+1)
     
      a = []
      for i in range(n):
          val = s - b[i] + b[(i-1)%n]
          if val % n != 0 or val <= 0:
              a = []
              print ("NO")
              break
  
          a.append(str(val//n))
  
      if len(a) > 0:
          print ("YES")
          print (" ".join(a))
          
  