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 collections import defaultdict, deque, Counter
  from heapq import heapify, heappop, heappush
  import math
  from copy import deepcopy
  from itertools import combinations, permutations, product, combinations_with_replacement
  from bisect import bisect_left, bisect_right
  
  import sys
  
  def input():
      return sys.stdin.readline().rstrip()
  def getN():
      return int(input())
  def getNM():
      return map(int, input().split())
  def getList():
      return list(map(int, input().split()))
  def getListGraph():
      return list(map(lambda x:int(x) - 1, input().split()))
  def getArray(intn):
      return [int(input()) for i in range(intn)]
  
  mod = 10 ** 9 + 7
  MOD = 998244353
  # sys.setrecursionlimit(10000000)
  inf = float('inf')
  eps = 10 ** (-15)
  dy = [0, 1, 0, -1]
  dx = [1, 0, -1, 0]
  
  #############
  # Main Code #
  #############
  
  T = getN()
  for _ in range(T):
      N = getN()
      A = getList()
  
      L1 = [0] * N
      for i in range(N):
          L1[i] = A[i] - A[i - 1]
  
      # L2[i] = n * (a[i-1] - a[i])
      # 全部0ならA[0]の約数でやる
      L2 = [0] * N
      for i in range(N):
          L2[i] = L1[i] - L1[i - 1]
  
      # 全部がNの倍数か
      flag = True
      for i in range(N):
          if abs(L2[i]) % N != 0:
              flag = False
              break
          L2[i] //= N
  
      if not flag:
          print('NO')
          continue
  
      # a1との差を表す
      for i in range(1, N):
          L2[i] += L2[i - 1]
  
      # あとは二分探索してピタリが出るか
      def f(x):
          res = 0
          for i in range(N):
              res += (x - L2[-i]) * (i + 1)
          return res
  
      ok, ng = max(1, max(L2) + 1), 10 ** 18
      while abs(ng - ok) > 1:
          mid = (ok + ng) // 2
          if f(mid) <= A[0]:
              ok = mid
          else:
              ng = mid
  
      opt = f(ok)
      if opt == A[0]:
          print('YES')
          print(*[ok - l for l in L2])
      else:
          print('NO')