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 CPP):

Solution:
#include <bits/stdc++.h> 
#include<cmath>
#include<string>
#define pb push_back
#define x first
#define y second
#define fast ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define all(x) x.begin(),x.end()
//#define int long long
#define ll long long
using namespace std;
 
void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
//void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}
 
template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifndef ONLINE_JUDGE
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif
 
// look for  all edge cases
//search for a pattern


void solve(){
	int n;
	cin >> n;
	vector<int> v(n);
	for(int i =0 ; i< n; i++){
		cin >> v[i];
	}
	
	vector<int> dp(n+1, 0);
	
	for(int i =0 ; i < n; i++){
		if(v[i] == 0)
			continue;
			
		int ma = max(0,i - (v[i] - 1));
		dp[ma] += 1;
		dp[i+1] += -1;
		
	}
	for(int i = 1; i < n; i++){
		dp[i] = dp[i] + dp[i-1];
	}
	
	for(int i =0 ; i< n; i++){
		if(dp[i] <= 0){
			cout << 0 << " ";
		}
		else
			cout << 1 << " ";
	}
	cout << "\n";
	
}

signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);	
		
	int t;
	cin >> t;
	while(t--){
		solve();
	}
		
			
	return 0;
	
}