human
Can you provide 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:
.

# SHRi GANESHA    author: Kunal Verma #

import os
import random
import sys
from collections import Counter
from heapq import heapify, heappop

from itertools import *

from io import BytesIO, IOBase
from math import inf, gcd

def lcm(a, b):
    return (a * b) // gcd(a, b)

'''
    mod = 10 ** 9 + 7
    fac = [1]
    for i in range(1, 2 * 10 ** 5 + 1):
        fac.append((fac[-1] * i) % mod)
    fac_in = [pow(fac[-1], mod - 2, mod)]
    for i in range(2 * 10 ** 5, 0, -1):
        fac_in.append((fac_in[-1] * i) % mod)
    fac_in.reverse()
    def comb(a, b):
        if a < b:
            return 0
        return (fac[a] * fac_in[b] * fac_in[a - b]) % mod
'''

# MAXN = 10000004
# spf = [0 for i in range(MAXN)]
# adj = [[] for i in range(MAXN)]
def sieve():
    global spf, adj, MAXN
    spf[1] = 1
    for i in range(2, MAXN):
        spf[i] = i

    for i in range(2, MAXN):
        if i * i > MAXN:
            break
        if (spf[i] == i):
            for j in range(i * i, MAXN, i):
                if (spf[j] == j):
                    spf[j] = i

def getdistinctFactorization(n):
    global adj, spf, MAXN
    for i in range(1, n + 1):
        index = 1
        x = i
        if (x != 1):
            adj[i].append(spf[x])
        x = x // spf[x]
        while (x != 1):
            if (adj[i][index - 1] != spf[x]):
                adj[i].append(spf[x])
                index += 1
            x = x // spf[x]

def printDivisors(n):
    i = 2
    z = [1, n]
    while i <= sqrt(n):
        if (n % i == 0):
            if (n / i == i):
                z.append(i)
            else:
                z.append(i)
                z.append(n // i)
        i = i + 1
    return z

def create(n, x, f):
    pq = len(bin(n)[2:])
    if f == 0:
        tt = min
    else:
        tt = max
    dp = [[inf] * n for _ in range(pq)]
    dp[0] = x
    for i in range(1, pq):
        for j in range(n - (1 << i) + 1):
            dp[i][j] = tt(dp[i - 1][j], dp[i - 1][j + (1 << (i - 1))])
    return dp

def enquiry(l, r, dp, f):
    if...