Problem: This week Arkady wanted to cook some pancakes (to follow ancient traditions) and make a problem about that. But then he remembered that one can't make a problem about stacking pancakes without working at a specific IT company, so he decided to bake the Napoleon cake instead.

To bake a Napoleon cake, one has to bake n dry layers first, and then put them on each other in one stack, adding some cream. Arkady started with an empty plate, and performed the following steps n times: 

  * place a new cake layer on the top of the stack; 
  * after the i-th layer is placed, pour a_i units of cream on top of the stack. 



When x units of cream are poured on the top of the stack, top x layers of the cake get drenched in the cream. If there are less than x layers, all layers get drenched and the rest of the cream is wasted. If x = 0, no layer gets drenched.

<image> The picture represents the first test case of the example.

Help Arkady determine which layers of the cake eventually get drenched when the process is over, and which don't.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 20 000). Description of the test cases follows.

The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of layers in the cake.

The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ n) — the amount of cream poured on the cake after adding each layer.

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case, print a single line with n integers. The i-th of the integers should be equal to 1 if the i-th layer from the bottom gets drenched, and 0 otherwise.

Example

Input


3
6
0 3 0 0 1 3
10
0 0 0 1 0 5 0 0 0 2
3
0 0 0


Output


1 1 0 1 1 1 
0 1 1 1 1 1 0 0 1 1 
0 0 0 

Solution (in PYTHON3):

Solution:
from collections import defaultdict,OrderedDict,Counter
from sys import stdin,stdout
from bisect import bisect_left,bisect_right
# import numpy as np
from queue import Queue,PriorityQueue
from heapq import *
from statistics import *
from math import *
import fractions
import copy
from copy import deepcopy
import sys
import io
sys.setrecursionlimit(10000)
import math
import os
import bisect
import collections
mod=pow(10,9)+7
import random
from random import *
from time import  time;
def ncr(n, r, p=mod):
    num = den = 1
    for i in range(r):
        num = (num * (n - i)) % p
        den = (den * (i + 1)) % p
    return (num * pow(den,
            p - 2, p)) % p
def normalncr(n,r):
    r=min(r,n-r)
    count=1;
    for i in range(n-r,n+1):
        count*=i;

    for i in range(1,r+1):
        count//=i;
    return count

inf=float("inf")
adj=defaultdict(set)
visited=defaultdict(int)
def addedge(a,b):
    adj[a].add(b)
    adj[b].add(a)
def bfs(v):
    q=Queue()
    q.put(v)
    visited[v]=1
    while q.qsize()>0:
        s=q.get_nowait()
        print(s)
        for i in adj[s]:
            if visited[i]==0:
                q.put(i)
                visited[i]=1
def dfs(v,visited):
    if visited[v]==1:
        return;
    visited[v]=1

    print(v)
    for i in adj[v]:
        dfs(i,visited)
# a9=pow(10,6)+10
# prime = [True for i in range(a9 + 1)]
# def SieveOfEratosthenes(n):
#     p = 2
#     while (p * p <= n):
#         if (prime[p] == True):
#             for i in range(p * p, n + 1, p):
#                 prime[i] = False
#         p += 1
# SieveOfEratosthenes(a9)
# prime_number=[]
# for i in range(2,a9):
#     if prime[i]:
#         prime_number.append(i)
def reverse_bisect_right(a, x, lo=0, hi=None):
    if lo < 0:
        raise ValueError('lo must be non-negative')
    if hi is None:
        hi = len(a)
    while lo < hi:
        mid = (lo+hi)//2
        if x > a[mid]:
            hi = mid
        else:
            lo = mid+1
    return lo
def reverse_bisect_left(a, x, lo=0, hi=None):
    if lo < 0:
        raise ValueError('lo must be non-negative')
    if hi is None:
        hi = len(a)
    while lo < hi:
        mid = (lo+hi)//2
        if x >= a[mid]:
            hi = mid
        else:
            lo = mid+1
    return lo

def get_list():
    return list(map(int,input().split()))
def get_str_list_in_int():
    return [int(i) for i in list(input())]
def get_str_list():
    return list(input())
def get_map():
    return map(int,input().split())
def input_int():
    return int(input())
def matrix(a,b):
    return [[0 for i in range(b)] for j in range(a)]
def swap(a,b):
    return b,a
def find_gcd(l):
    a=l[0]
    for i in range(len(l)):
        a=gcd(a,l[i])
    return a;
def is_prime(n):
    sqrta=int(sqrt(n))
    for i in range(2,sqrta+1):
        if n%i==0:
            return 0;
    return 1;

def prime_factors(n):

    sqrta = int(sqrt(n))
    for i in range(2,sqrta+1):
        if n%i==0:
            return [i]+prime_factors(n//i)
    return [n]
def p(a):
    if type(a)==str:
        print(a+"\n")
    else:
        print(str(a)+"\n")
def ps(a):
    if type(a)==str:
        print(a)
    else:
        print(str(a))
def kth_no_not_div_by_n(n,k):
    return k+(k-1)//(n-1)
def forward_array(l):
    n=len(l)
    stack = []
    forward=[0]*n
    for i in range(len(l) - 1, -1, -1):
        while len(stack) and l[stack[-1]] < l[i]:
            stack.pop()
        if len(stack) == 0:
            forward[i] = len(l);
        else:
            forward[i] = stack[-1]
        stack.append(i)
    return forward;
def backward_array(l):
    n=len(l)
    stack = []
    backward=[0]*n
    for i in range(len(l)):
        while len(stack) and l[stack[-1]] < l[i]:
            stack.pop()
        if len(stack) == 0:
            backward[i] = -1;
        else:
            backward[i] = stack[-1]
        stack.append(i)
    return backward
nc="NO"
yc="YES"
ns="No"
ys="Yes"

input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# input=stdin.readline
# print=stdout.write

t=int(input())
for i in range(t):
    n=int(input())
    l=get_list()
    if max(l)==1:
        print(*l)
        continue
    l.reverse()
    m=defaultdict(int)
    for i in range(n):
        m[i]+=1
        m[i+l[i]]-=1

        m[i]+=m[i-1]
    k=[0]*n
    for i in range(n):
        if m[i]:
            k[i]=1
    k.reverse()
    print(*k)