Problem: A permutation — is a sequence of length n integers from 1 to n, in which all the numbers occur exactly once. For example, [1], [3, 5, 2, 1, 4], [1, 3, 2] — permutations, and [2, 3, 2], [4, 3, 1], [0] — no.

Polycarp was recently gifted a permutation a[1 ... n] of length n. Polycarp likes trees more than permutations, so he wants to transform permutation a into a rooted binary tree. He transforms an array of different integers into a tree as follows: 

  * the maximum element of the array becomes the root of the tree; 
  * all elements to the left of the maximum — form a left subtree (which is built according to the same rules but applied to the left part of the array), but if there are no elements to the left of the maximum, then the root has no left child; 
  * all elements to the right of the maximum — form a right subtree (which is built according to the same rules but applied to the right side of the array), but if there are no elements to the right of the maximum, then the root has no right child. 



For example, if he builds a tree by permutation a=[3, 5, 2, 1, 4], then the root will be the element a_2=5, and the left subtree will be the tree that will be built for the subarray a[1 ... 1] = [3], and the right one — for the subarray a[3 ... 5] = [2, 1, 4]. As a result, the following tree will be built: 

<image> The tree corresponding to the permutation a=[3, 5, 2, 1, 4].

Another example: let the permutation be a=[1, 3, 2, 7, 5, 6, 4]. In this case, the tree looks like this: 

<image> The tree corresponding to the permutation a=[1, 3, 2, 7, 5, 6, 4].

Let us denote by d_v the depth of the vertex a_v, that is, the number of edges on the path from the root to the vertex numbered a_v. Note that the root depth is zero. Given the permutation a, for each vertex, find the value of d_v.

Input

The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.

The first line of each test case contains an integer n (1 ≤ n ≤ 100) — the length of the permutation.

This is followed by n numbers a_1, a_2, …, a_n — permutation a.

Output

For each test case, output n values — d_1, d_2, …, d_n.

Example

Input


3
5
3 5 2 1 4
1
1
4
4 3 1 2


Output


1 0 2 3 1 
0 
0 1 3 2 

Solution (in CPP):

Solution:
// Jai Mata Di..
// 
// @ author : Sanjit Anand (sanjit_15)
// 
#include <bits/stdc++.h>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;

#define int             long long
#define PI              3.1415926535897932384626
#define gc              getchar_unlocked
#define fo(i,n)         for(i=0;i<n;i++)
#define rep(i,k,n)      for(i=k;k<n?i<n:i>n;k<n?i+=1:i-=1)
#define deb(x)          cout << #x << "=" << x << endl
#define deb2(x, y)      cout << #x << "=" << x << "," << #y << "=" << y << endl
#define endl            "\n"
#define pb              push_back
#define pf              pop_front
#define ppb             pop_back
#define ppf             pop_front
#define mp              make_pair
#define F               first
#define S               second
#define trav(it, a)     for(auto it = a.begin(); it != a.end(); it++)
#define ppc             __builtin_popcount
#define ppcll           __builtin_popcountll

template<typename T1,typename T2>istream& operator>>(istream& in,pair<T1,T2> &a){in>>a.fr>>a.sc;return in;}
template<typename T1,typename T2>ostream& operator<<(ostream& out,pair<T1,T2> a){out<<a.fr<<" "<<a.sc;return out;}
template<typename T,typename T1>T amax(T &a,T1 b){if(b>a)a=b;return a;}
template<typename T,typename T1>T amin(T &a,T1 b){if(b<a)a=b;return a;}

typedef pair<int, int>  pii;
typedef vector<int>     vi;
typedef vector<pii>     vpii;
typedef vector<vi>      vvi;
typedef set<int>        si;
typedef list<int>       li;

const long long INF = 1e18;
const long long MOD = 1000000007;
const long long MM = 998244353;

const int N = 10005;

int mpow(int base, int exp, int m);
int gcd(int a, int b);



void rec(int ar[], int start, int end, int level[], int t){

	if(start >= end){
		level[start]=t;
		return;
	}
	
	int max_element = -1*INF, max_idx = -1;

	for(int i=start; i<=end; i++){
		if(ar[i] > max_element){
			max_element = ar[i];
			max_idx = i;
		}
	}
	// cout << max_idx << "\n";
	level[max_idx]=t;
	t++;
	if(start <= max_idx-1){
		rec(ar,start,max_idx-1,level,t);
	}
	if(end >= max_idx+1){
		rec(ar,max_idx+1,end,level,t);
	}

	return;

}


void solve() {
    int n;
    cin >> n;
    int level[n];
    int ar[n];
    for(int i=0; i<n; i++){
    	cin >> ar[i];
    	level[i] = 0;
    }

    rec(ar,0,n-1,level,0);
    for(int i=0; i<n; i++){
    	cout << level[i] << " ";
    }
    cout << "\n";



}

signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);

    #ifdef NCR
        pmain();
    #endif
    #ifdef SIEVE
        sieve();
    #endif

    int t = 1;
    cin >> t;
    while(t--) {
      solve();
    }

    return 0;
}

int gcd(int a, int b){
    return b==0 ? a : gcd(b,a%b);
}

int mpow(int base, int exp, int m=MOD) {
  base %= m;
  int result = 1;
  while (exp > 0) {
    if (exp & 1) result = ((long long)result * base) % m;
    base = ((long long)base * base) % m;
    exp >>= 1;
  }
  return result;
}