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 JAVA):

Solution:


import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class experiment
{
    static int M = 1_000_000_007;
    static int INF = Integer.MAX_VALUE;
    static final FastScanner fs = new FastScanner();

    //variable
    static int[][] arr;
    public static void main(String[] args) throws IOException
    {
        int T = fs.nextInt();

        outer : while(T-->0) {
            int n = fs.nextInt();
            arr = new int[n][2];

            for(int i=0; i<n; i++) {
                arr[i][0] = fs.nextInt();
                arr[i][1] = -1;
            }
            setDepth(0,n-1,0);
           for(int i=0; i<n; i++) {
               System.out.print(arr[i][1]+" ");
           }
           System.out.println();
        }

    }
    //function
    static void setDepth(int i, int j,int depth) {
        int index = maxIndex(i,j);
        if(index != -1) {
            arr[index][1] = depth;
            setDepth(i, index-1,depth+1);
            setDepth(index+1,j,depth+1);
        }

    }
    static int maxIndex(int i, int j) {
        int max = -INF;
        int index = -1;
        for(;i<=j;i++){
            if(max<arr[i][0]) {
                max =arr[i][0];
                index = i;
            }
        }
        return index;
    }
    // Template
    static long nCk(int n, int k) {
        return (modMult(fact(n),fastexp(modMult(fact(n-k),fact(k)),M-2)));
    }
    static long fact (long n) {

        long fact =1;
        for(int i=1; i<=n; i++) {
            fact = modMult(fact,i);
        }
        return fact%M;
    }
    static long modMult(long a,long b) {
        return a*b%M;
    }
    static long fastexp(long x, int y){
        if(y==1) return x;

        long ans = fastexp(x,y/2);
        if(y%2 == 0) return modMult(ans,ans);
        else return modMult(ans,modMult(ans,x));
    }
    static final Random random = new Random();

    static void ruffleSort(int arr[])
    {
        int n = arr.length;
        for(int i=0; i<n; i++)
        {
            int j = random.nextInt(n),temp = arr[j];
            arr[j] = arr[i];
            arr[i] = temp;
        }
        Arrays.sort(arr);
    }


    public static class Pairs implements Comparable<Pairs>
    {
        int value,index;
        Pairs(int value, int index)
        {
            this.value = value;
            this.index = index;
        }
        public int compareTo(Pairs p)
        {
            return Integer.compare(this.value, p.value);
        }
    }
}
class FastScanner
{
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer str = new StringTokenizer("");

    String next() throws IOException
    {
        while(!str.hasMoreTokens())
            str = new StringTokenizer(br.readLine());

        return str.nextToken();
    }

    char nextChar() throws IOException {
        return next().charAt(0);
    }

    int nextInt() throws IOException
    {
        return Integer.parseInt(next());
    }

    float nextfloat() throws IOException
    {
        return Float.parseFloat(next());
    }

    double nextDouble() throws IOException
    {
        return Double.parseDouble(next());
    }
    long nextLong() throws IOException
    {
        return Long.parseLong(next());
    }
    byte nextByte() throws IOException
    {
        return Byte.parseByte(next());
    }
    int [] arrayIn(int n) throws IOException
    {
        int  arr[] = new int[n];
        for(int i=0; i<n; i++)
        {
            arr[i] = nextInt();
        }
        return arr;
    }
}