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:
//firstly save by ctrl+s before running the code
//press f5 to debug and input in terminal
//typcast by e.g (long long)variable and for constant e.g 5ll
//-----------------------------------------------------------------------------------------------------------------------------
//count set bits using __builtin_popcount(n);
// cout << fixed << setprecision(10) << pi <<" "<<npi<<endl;-->synax to set precision for float/double datatype pi , npi
//ceil or floor used for double and return double 
//log also return double
//use floor(n)==n to see if a double n is integer
//for int x,y to get ceil in x/y do as (x+(y-1))/y and x/y itself equals floor only (but returns int)
//log2(n)==floor(log2(n))-->to check if power of 2 
//------------------------------------------FUN BEIGINS!-----------------------------------------------------------------------
#include<bits/stdc++.h>
#include<fstream>
#define N 1000000007
#define M 998244353
using namespace std;
#define ll long long
#define pll  pair<ll,ll>
#define vll  vector<ll>
#define vpll  vector<pll>
#define  vvll vector<vll>
#define endl "\n"                   //"\n" not flushes output but endl do
#define umap unordered_map<ll,ll>
//------------------------------------------------------------------------------------------------------------------------------
vector<vector<long long>>adj;
map<long,bool>vis,viss; 
vector<long long>rnk,parent,sz;
#define sieve_max 1000000
ll spf[sieve_max+1];
umap mp;
void yg(vll v, ll l, ll r,ll prev){
    if(r<l)return;

    if(r==l){mp[r]=prev; return;}
    ll mx=LLONG_MIN;
    ll ct=0;

  for(ll i=l;i<=r;i++){if(v[i]>mx){
      mx=v[i]; ct=i;
  }}
        
    mp[ct]=prev;

    yg(v,l,ct-1,prev+1); yg(v,ct+1,r,prev+1);

    return;


}

int main(){
    ios_base::sync_with_stdio(false);    cin.tie(NULL);
    ll t,temp;
    cin>>t; 
    while(t--){
        mp.clear();
        ll n;
        cin>>n; vll v;
        for(ll i=0;i<n;i++){
            cin>>temp; v.push_back(temp);
        }
        yg(v,0,n-1,0);

        for(ll i=0;i<n;i++)cout<<mp[i]<<" ";
        cout<<endl;


    }
   

       
     
return 0;
}