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: # cook your dish here
  import sys
  #import random
  from bisect import bisect_left as lb
  from bisect import bisect_right as rb
  from collections import deque
  #sys.setrecursionlimit(10**8)
  from queue import PriorityQueue as pq
  from math import gcd
  #import math
  input_ = lambda: sys.stdin.readline().strip("\r\n")
  ii = lambda : int(input_())
  il = lambda : list(map(int, input_().split()))
  ilf = lambda : list(map(float, input_().split()))
  lii = lambda : list(map(int, list(ip())))
  ip = lambda : input_()
  fi = lambda : float(input_())
  ap = lambda ab,bc,cd : ab[bc].append(cd)
  li = lambda : list(input_())
  pr = lambda x : print(x)
  prinT = lambda x : print(x)
  f = lambda : sys.stdout.flush()
  inv =lambda x:pow(x,mod-2,mod)
  dx = [0,0,1,-1]
  dy = [1,-1,0,0]
  mod = 10**9 + 7
  mod1 = 998244353
  
  for _ in range (ii()) :
      n = ii()
      a = il()
  
      s = sum(a)
  
      x = (n*(n+1))//2
  
      if (s%x == 0 and s//x >= n) :
          
          ans = []
  
          t = s//x
          i = n-1
          fl = 0
          for k in range (n) :
              y = a[i] + t
  
              t1 = a[(i+1)%n]
  
              z = y - t1
  
              if (z%n != 0) :
                  fl = 1
                  break
  
              ans.append(z//n)
              if (ans[-1] <= 0 or ans[-1] >10**9) :
                  fl = 1
                  break
              i += 1
              i %= n
          if (fl) :
              print("NO")
              continue
          print("YES")
          print(*ans)
          
      else :
          print("NO")
  