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. 
from sys import stdin,stdout,setrecursionlimit
from math import gcd,sqrt,factorial,pi,inf
from collections import deque,defaultdict
from bisect import bisect,bisect_left
from time import time
from itertools import permutations as per
from heapq import heapify,heappush,heappop,heappushpop
input=stdin.readline
R=lambda:map(int,input().split())
I=lambda:int(input())
S=lambda:input().rstrip('\r\n')
L=lambda:list(R())
P=lambda x:stdout.write(str(x)+'\n')
lcm=lambda x,y:(x*y)//gcd(x,y)
nCr=lambda x,y:(f[x]*inv((f[y]*f[x-y])%N))%N
inv=lambda x:pow(x,N-2,N)
sumx=lambda x:(x**2+x)//2
N=10**9+7

for _ in range(I()):
    n=I()
    a=L()
    if n==1:
        print('YES')
        print(*a)
        continue
    c=sumx(n)
    s=sum(a)
    if s%c:
        print('NO')
        continue
    s=s//c
    ans=[]
    f=True
    for i in range(n):
        #print((a[i-1]-(a[i]-s))%n,a[i-1]>a[i]-s,s<a[i])
        if s<a[i] and a[i-1]>a[i]-s and not (a[i-1]-(a[i]-s))%n:
            ans+=(a[i-1]-(a[i]-s))//n,
        else:
            f=False
            break
    if f:
        print('YES')
        print(*ans)
    else:
        print('NO')



















