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 
///...................SUBHASHIS MOLLICK...................///
///.....DEPARTMENT OF COMPUTER SCIENCE AND ENGINEERING....///
///.............ISLAMIC UNIVERSITY,BANGLADESH.............///
///....................SESSION-(14-15)....................///
#include<bits/stdc++.h>
using namespace std;
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
#define sf(a) scanf("%d",&a)
#define sf2(a,b) scanf("%d %d",&a,&b)
#define sf3(a,b,c) scanf("%d %d %d",&a,&b,&c)
#define pf(a) printf("%d",a)
#define pf2(a,b) printf("%d %d",a,b)
#define pf3(a,b,c) printf("%d %d %d",a,b,c)
#define sfl(a) scanf("%lld",&a)
#define sfl2(a,b) scanf("%lld %lld",&a,&b)
#define sfl3(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define pfl(a) printf("%lld",a)
#define pfl2(a,b) printf("%lld %lld",a,b)
#define pfl3(a,b,c) printf("%lld %lld %lld",a,b,c)
#define nl printf("\n")
#define   timesave              ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define ll long long
#define pb push_back
#define MPI map<int,int>mp;
#define fr(i,n) for(i=0;i<n;i++)
#define fr1(i,n) for(i=1;i<=n;i++)
#define frl(i,a,b) for(i=a;i<=b;i++)
#define tz 100000
#define clr0(a) memset(a,0,sizeof(a))
#define clr1(a) memset(a,-1,sizeof(a))
#define space " "
#define yesp cout<<"YES"<<endl;
#define nop cout<<"NO"<<endl;
/*primes in range 1 - n
1 - 100(1e2) -> 25 pimes
1 - 1000(1e3) -> 168 primes
1 - 10000(1e4) -> 1229 primes
1 - 100000(1e5) -> 9592 primes
1 - 1000000(1e6) -> 78498 primes
1 - 10000000(1e7) -> 664579 primes
large primes ->
104729 1299709 15485863 179424673 2147483647 32416190071 112272535095293 48112959837082048697
*/
//freopen("Input.txt","r",stdin);
//freopen("Output.txt","w",stdout);
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
int dist[105];
vector<int>vec[105];
void dfs(int src,int par)
{
    if(vec[src].size()==0)
    {
        return;
    }
    for(int i=0; i<vec[src].size(); i++)
    {
        int v=vec[src][i];
        if(v!=par)
        {
            dist[v]=dist[src]+1;
            dfs(v,src);
        }
    }
}
main()
{
    timesave;
    int ts=1;
    cin>>ts;
    while(ts--)
    {
        int n,i,j;
        cin>>n;
        int ar[n+5]= {0},vis[105]={0};
        for(i=1; i<=n; i++)
        {
            cin>>ar[i];
        }
        for(i=1; i<=n; i++)
        {
            int mx=0;
            for(j=i-1; j>=1; j--)
            {
                if(ar[i]>ar[j] && vis[ar[j]]==0)
                {
                    mx=max(mx,ar[j]);
                }
            }
            if(mx)
            {
                vis[mx]=1;
                vec[ar[i]].push_back(mx);
            }
            mx=0;
            for(j=i+1; j<=n; j++)
            {
                if(ar[i]>ar[j] && vis[ar[j]]==0)
                {
                    mx=max(mx,ar[j]);
                }
            }
            if(mx)
            {
                vis[mx]=1;
                vec[ar[i]].push_back(mx);
            }
        }
        dfs(n,-1);
        for(i=1; i<=n; i++)
        {
            cout<<dist[ar[i]]<<" ";
            vec[ar[i]].clear();
            dist[ar[i]]=0;
        }
        cout<<endl;
    }
}


