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. 
# 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")
