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: t = int(input())
  for _ in range(t):
      n = int(input())
      B = list(map(int,input().split()))
      nsum = n * (n+1) / 2
      if sum(B) % nsum == 0:
          S = sum(B) // nsum
          ans = list()
          ll = 0
          for i in range(n):
              if i == 0:
                  kk = B[n-1]
              else:
                  kk = B[i-1]
              k = ((S-B[i]+kk) / n)
              if k < 1 or k != int(k):
                  ll = 1
                  break
              ans.append(k)
          if ll == 1:
              print('NO')
          else:
              print('YES')
              for i in ans:
                  print(int(i),end=' ')
      else:
          print('NO')
  
  
  
  
  
  
  
  
  
  # import numpy as np
  # def eliminate(r1, r2, col, target=0):
  #     fac = (r2[col]-target) / r1[col]
  #     for i in range(len(r2)):
  #         r2[i] -= fac * r1[i]
  
  # def gauss(a):
  #     for i in range(len(a)):
  #         if a[i][i] == 0:
  #             for j in range(i+1, len(a)):
  #                 if a[i][j] != 0:
  #                     a[i], a[j] = a[j], a[i]
  #                     break
  #             else:
  #                 raise ValueError("Matrix is not invertible")
  #         for j in range(i+1, len(a)):
  #             eliminate(a[i], a[j], i)
  #     for i in range(len(a)-1, -1, -1):
  #         for j in range(i-1, -1, -1):
  #             eliminate(a[i], a[j], i)
  #     for i in range(len(a)):
  #         eliminate(a[i], a[i], i, target=1)
  #     return a
  
  # def inverse(a):
  #     tmp = [[] for _ in a]
  #     for i,row in enumerate(a):
  #         assert len(row) == len(a)
  #         tmp[i].extend(row + [0]*i + [1] + [0]*(len(a)-i-1))
  #     gauss(tmp)
  #     ret = []
  #     for i in range(len(tmp)):
  #         ret.append(tmp[i][len(tmp[i])//2:])
  #     return ret
  
  # t = int(input())
  
  # for _ in range(t):
  #     n = int(input())
  #     b = list(map(int,input().split()))
  #     B = np.array([[b[i]] for i in range(n)])
  #     a = [1]
  #     for i in range(n, 1, -1):
  #         a.append(i)
  #     A = list()
  #     A.append(a)
  #     for i in range(1, n):
  #         temp = A[i-1][0:n-1]
  #         temp = [A[i-1][-1]] + temp
  #         A.append(temp)
  #     Ar = np.array(A)
  #     # invA = np.linalg.inv(Ar)
  #     # print(invA)
  #     matrix = A
  #     inn = inverse(A)
  #     # print(inn)
  #     # invA = getMatrixInverse(A)
  #     # print(list(invA))
  #     invAA = np.array(inn)
  #     mat1, mat2 = list(invAA), list(B)
  #     result = [[0] for i in range(n)]
  #     for x in range(len(mat1)):
  #        # iterate through columns of c2
  #        for y in range(len(mat2[0])):
  #            # iterate through rows of r2
  #            for z in range(len(mat2)):
  #                 result[x][y] += mat1[x][z] * mat2[z][y]
  #     C = result
  #     kk = 0
  #     for i in range(n):
  #         if int(round(C[i][0])) < 1:
  #             kk = 1
  #     if kk == 1:
  #         print('NO')
  #     else:
  #         print('YES')
  #         for i in range(n):
  #             print(int(round(C[i][0])), end=' ')
  #         print()
  