human
Give me a Python program to print the answer as follows: given the following:
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 —...
Solve the task in PYTHON3.
assistant
#Here is Python program to print the answer as follows:
.

from collections import defaultdict, deque, Counter
from heapq import heapify, heappop, heappush
import math
from copy import deepcopy
from itertools import combinations, permutations, product, combinations_with_replacement
from bisect import bisect_left, bisect_right

import sys

def input():
    return sys.stdin.readline().rstrip()
def getN():
    return int(input())
def getNM():
    return map(int, input().split())
def getList():
    return list(map(int, input().split()))
def getListGraph():
    return list(map(lambda x:int(x) - 1, input().split()))
def getArray(intn):
    return [int(input()) for i in range(intn)]

mod = 10 ** 9 + 7
MOD = 998244353
# sys.setrecursionlimit(10000000)
inf = float('inf')
eps = 10 ** (-15)
dy = [0, 1, 0, -1]
dx = [1, 0, -1, 0]

#############
# Main Code #
#############

"""
構成するようなAを作れるか
a1, a2, a3...とする
b1 = a1 + 2 * an-1 + 3 * an-2...
b2 = a2 + 2 * a1 + 3 * an-1...
...

なる配列Aを作れるか

b2 - b1 = a1 + an-1 + an-2 +...+ a3 - (n-1)a2 = sum(A) - n * a2
b3 - b2 = sum(A) - n * a3
"""

T = getN()
for _ in range(T):
    N = getN()
    A = getList()

    L1 = [0] * N
    for i in range(N):
        L1[i] = A[i] - A[i - 1]

    # L2[i] = n * (a[i-1] - a[i])
    # 全部0ならA[0]の約数でやる
    L2 = [0] * N
    for i in range(N):
        L2[i] = L1[i] - L1[i - 1]

    # 全部がNの倍数か
    flag = True
    for i in range(N):
        if abs(L2[i]) % N != 0:
            flag = False
            break
        L2[i] //= N

    if not flag:
        print('NO')
        continue

    # a1との差を表す
    for i in range(1, N):
        L2[i] += L2[i - 1]

    # あとは二分探索してピタリが出るか
    def f(x):
        res = 0
        for i in range(N):
            res += (x - L2[-i]) * (i + 1)
        return res

    ok, ng = max(1, max(L2) + 1), 10 ** 18
    while abs(ng - ok) > 1:
        mid = (ok + ng) // 2
        if f(mid) <= A[0]:
            ok = mid
        else:
            ng = mid

    opt = f(ok)
    if opt == A[0]:
        print('YES')
        print(*[ok - l for l in L2])
    else:
        print('NO')