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: # SHRi GANESHA    author: Kunal Verma #
  
  import os
  import random
  import sys
  from collections import Counter
  from heapq import heapify, heappop
  
  from itertools import *
  
  from io import BytesIO, IOBase
  from math import inf, gcd
  
  
  def lcm(a, b):
      return (a * b) // gcd(a, b)
  
  
  '''
      mod = 10 ** 9 + 7
      fac = [1]
      for i in range(1, 2 * 10 ** 5 + 1):
          fac.append((fac[-1] * i) % mod)
      fac_in = [pow(fac[-1], mod - 2, mod)]
      for i in range(2 * 10 ** 5, 0, -1):
          fac_in.append((fac_in[-1] * i) % mod)
      fac_in.reverse()
      def comb(a, b):
          if a < b:
              return 0
          return (fac[a] * fac_in[b] * fac_in[a - b]) % mod
  '''
  
  
  # MAXN = 10000004
  # spf = [0 for i in range(MAXN)]
  # adj = [[] for i in range(MAXN)]
  def sieve():
      global spf, adj, MAXN
      spf[1] = 1
      for i in range(2, MAXN):
          spf[i] = i
  
      for i in range(2, MAXN):
          if i * i > MAXN:
              break
          if (spf[i] == i):
              for j in range(i * i, MAXN, i):
                  if (spf[j] == j):
                      spf[j] = i
  
  
  def getdistinctFactorization(n):
      global adj, spf, MAXN
      for i in range(1, n + 1):
          index = 1
          x = i
          if (x != 1):
              adj[i].append(spf[x])
          x = x // spf[x]
          while (x != 1):
              if (adj[i][index - 1] != spf[x]):
                  adj[i].append(spf[x])
                  index += 1
              x = x // spf[x]
  
  
  def printDivisors(n):
      i = 2
      z = [1, n]
      while i <= sqrt(n):
          if (n % i == 0):
              if (n / i == i):
                  z.append(i)
              else:
                  z.append(i)
                  z.append(n // i)
          i = i + 1
      return z
  
  
  def create(n, x, f):
      pq = len(bin(n)[2:])
      if f == 0:
          tt = min
      else:
          tt = max
      dp = [[inf] * n for _ in range(pq)]
      dp[0] = x
      for i in range(1, pq):
          for j in range(n - (1 << i) + 1):
              dp[i][j] = tt(dp[i - 1][j], dp[i - 1][j + (1 << (i - 1))])
      return dp
  
  
  def enquiry(l, r, dp, f):
      if l > r:
          return inf if not f else -inf
      if f == 1:
          tt = max
      else:
          tt = min
      pq1 = len(bin(r - l + 1)[2:]) - 1
      return tt(dp[pq1][l], dp[pq1][r - (1 << pq1) + 1])
  
  
  def SieveOfEratosthenes(n):
      prime = [True for i in range(n + 1)]
      p = 2
      while (p * p <= n):
          if (prime[p] == True):
              for i in range(p * p, n + 1, p):
                  prime[i] = False
          p += 1
      x = []
      for i in range(2, n + 1):
          if prime[i]:
              x.append(i)
      return x
  
  
  
  def main():
  
      for _ in range(int(input())):
          n=int(input())
          b=[int(x) for x in input().split()]
          if sum(b)%((n*(n+1))//2)!=0:
              print("NO")
              continue
          s=sum(b)//((n*(n+1))//2)
          # print("sum",s)
          a=[]
          for i in range(n-1,-1,-1):
              a.append((s-b[i]+b[i-1])//n)
  
          for i in a:
  
              if i<=0 :
                  print("NO")
                  break
  
          else:
              if sum(a)!=s:
                  print("NO")
                  continue
  
              for i in range(n-1,-1,1):
                  if b[i]-s+n*a[i]!=b[i-1]:
                      print("NO")
                      break
              else:
                  print("YES")
                  print(*reversed(a))
  
  # Fast IO Region
  BUFSIZE = 8192
  
  
  class FastIO(IOBase):
      newlines = 0
  
      def __init__(self, file):
          self._fd = file.fileno()
          self.buffer = BytesIO()
          self.writable = "x" in file.mode or "r" not in file.mode
          self.write = self.buffer.write if self.writable else None
  
      def read(self):
          while True:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              if not b:
                  break
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines = 0
          return self.buffer.read()
  
      def readline(self):
          while self.newlines == 0:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              self.newlines = b.count(b"\n") + (not b)
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines -= 1
          return self.buffer.readline()
  
      def flush(self):
          if self.writable:
              os.write(self._fd, self.buffer.getvalue())
              self.buffer.truncate(0), self.buffer.seek(0)
  
  
  class IOWrapper(IOBase):
      def __init__(self, file):
          self.buffer = FastIO(file)
          self.flush = self.buffer.flush
          self.writable = self.buffer.writable
          self.write = lambda s: self.buffer.write(s.encode("ascii"))
          self.read = lambda: self.buffer.read().decode("ascii")
          self.readline = lambda: self.buffer.readline().decode("ascii")
  
  
  sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
  input = lambda: sys.stdin.readline().rstrip("\r\n")
  
  if __name__ == '__main__':
      main()