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:
/*
ID: shubham147
TASK: gift1
LANG: C++14                 
*/
// #include <ext/pb_ds/assoc_container.hpp> // Common file 
// #include <ext/pb_ds/tree_policy.hpp> 
// #include <functional> // for less
#include<bits/stdc++.h>
// #pragma GCC optimize("Ofast")
// #pragma GCC target("avx,avx2,fma")
// #pragma GCC optimization("unroll-loops")
// #pragma GCC optimize("unroll-loops")
// #pragma GCC optimize("fast-math")
// #pragma GCC optimize("no-stack-protector")
#define ll long long
#define f(i,a,b) for(ll i=a;i<b;i++)
#define mod 1000000007
#define mp make_pair
#define ff first
#define ss second
#define rf(i,a,b) for(ll i=a;i>=b;i--)
#define sc(a) scanf("%lld",&a)
#define pf printf
#define sz(a) (int)(a.size())
#define psf push_front
#define ppf pop_front
#define ppb pop_back
#define pb push_back
#define pq priority_queue
#define all(s) s.begin(),s.end()
#define sp(a) setprecision(a)
#define rz resize
#define ld long double
#define inf (ll)1e18
#define ub upper_bound
#define lb lower_bound
#define bs binary_search
#define eb emplace_back
const double pi = acos(-1);
ll binpow(ll a, ll b){ll res=1;while(b!=0){if(b&1)res*=a;a*=a;b>>=1;}return res;}
 
// using namespace __gnu_pbds; 
using namespace std;

// typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_nd_update> new_data_set;

ll n;
vector<ll> a,d;

void fn(ll l, ll r, ll depth)
{
    if(l>r)
        return;
    ll id=l,mx=-inf;
    f(i,l,r+1)
    {
        if(a[i]>mx)
            mx=a[i],id=i;
    }
    d[id]=depth;
    if(id!=l)
        fn(l,id-1,depth+1);
    if(id!=r)
        fn(id+1,r,depth+1);
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    // freopen("gift1.in","r",stdin);
    // freopen("gift1.out","w",stdout);
#ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
#endif
    ll t=1;
    cin>>t;
    while(t--)
    {
        cin>>n;
        a.rz(n),d.rz(n);
        f(i,0,n)
            cin>>a[i];
        fn(0LL,n-1,0LL);
        f(i,0,n)
            cout<<d[i]<<" ";
        cout<<"\n";
        a.clear(),d.clear();
    }
}