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:
/*
    JAI MATA DI
 */

import java.util.*;



import java.io.*;
import java.math.BigInteger;

public class CP {
	  static class FR{
			BufferedReader br;
			StringTokenizer st;
			public FR() {
				br = new BufferedReader(new InputStreamReader(System.in));
			}
			String next() 
		    { 
		        while (st == null || !st.hasMoreElements()) 
		        { 
		            try
		            { 
		                st = new StringTokenizer(br.readLine()); 
		            } 
		            catch (IOException  e) 
		            { 
		                e.printStackTrace(); 
		            } 
		        } 
		        return st.nextToken(); 
		    } 

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

		    long nextLong() 
		    { 
		        return Long.parseLong(next()); 
		    } 

		    double nextDouble() 
		    { 
		        return Double.parseDouble(next()); 
		    } 

		    String nextLine() 
		    { 
		        String str = ""; 
		        try
		        { 
		            str = br.readLine(); 
		        } 
		        catch (IOException e) 
		        { 
		            e.printStackTrace(); 
		        } 
		        return str; 
		    } 
		}
	 static int mod  = 1000000007;
	 static int gcd(int a, int b)
	  {      
	     if (b == 0)
	        return a;
	     return gcd(b, a % b); 
	  }
	 static boolean[] prime(int num) {
			boolean[] bool = new boolean[num];
		     
		      for (int i = 0; i< bool.length; i++) {
		         bool[i] = true;
		      }
		      for (int i = 2; i< Math.sqrt(num); i++) {
		         if(bool[i] == true) {
		            for(int j = (i*i); j<num; j = j+i) {
		               bool[j] = false;
		            }
		         }
		      }
		      if(num >= 0) {
		    	  bool[0] = false;
		    	  bool[1] = false;
		      }
		      
		      return bool;
		}
	 static class Pair implements Comparable<Pair>{
	 		int first;
	 		int second;
	 		Pair(int key , int value){
	 			this.first = key;
	 			this.second = value;
	 		}
			@Override
			public int compareTo(Pair o) {
				if(this.second == o.second) {
					return this.first - o.first;
				}
//				if(this.first == o.first) this.second - o.second;
				return -(this.second - o.second);
			}
			@Override
			public int hashCode(){
				return first + second;
			}
	 	}
	 	
	
	 public static void main(String args[]) {
		 
		 FR sc = new FR();
		 StringBuilder sb = new StringBuilder();
		 int t = sc.nextInt();
		 while(t-- > 0) {
			int n = sc.nextInt();
			int dp[] = new int[n+1]; 
			int[] arr = new int[n];
			for(int i = 0 ; i<n ;i++) arr[i] = sc.nextInt();
			Tree tree = new Tree(arr);
			 tree.depth(dp);
			 for(int i = 0 ; i<n; i++) {
				 sb.append(dp[arr[i]]+" ");
			 }
			 sb.append("\n");
		 }
		 System.out.println(sb);
		
	 
	 
	 }
	 
		}
	class Tree{
		class Node{
			int data;
			Node left;
			Node right;
			Node(int data){
				this.data  =data;
				left = right =null;
			}
		}
		private Node root = null;
		
		int find(int[] arr  , int start , int end){
			int max = -1;
			int c = -1;
			for(int i = start ; i<=end ;i++ ) {
				if(c < arr[i]) {
					max = i;
					c = arr[i];
				}
				
			}
			return max;
		}
		
		Tree(int[] arr){
			this.root = tree(arr, 0, arr.length-1);
		}
		Node tree(int[] arr , int s , int e){
			
			int pe = find(arr, s, e);
			if(pe == -1) return null;
			Node root = new Node(arr[pe]);
			root.left = tree(arr, s, pe-1);
			root.right= tree(arr, pe+1, e);
			
			return root;
		}
		
		
		 void depth(int[] dp) {
			d(this.root , dp);
		}
		 private void d(Node root , int[] dp ) {
			if(root == null) return ;
			int left = 0 , right = 0;
			if(root.left != null ) {
				dp[root.left.data] += dp[root.data] + 1;
				d(root.left, dp);
			}
			if(root.right != null) {
				dp[root.right.data] += dp[root.data] + 1;
				d(root.right ,dp);
			}
		}
	}