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.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.PriorityQueue;
import java.util.StringTokenizer;
import java.io.*;
 
public class PermutationTrans {
	
	private static class MyScanner {
	    BufferedReader br;
	    StringTokenizer st;

	    public MyScanner() {
	       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;
	    }

	 }
	
	public static class SegmentTrees { 
	      
	    static int n; // array size 
	    static long[] tree; // to keep track of the values
	    static int[] ind; //to keep track of the index  
		
	    SegmentTrees(int n)
	    {
	    	this.n = n;
	    	tree = new long[2*n];
	    	ind = new int[2*n];
	    }

	    // function to build the tree 
	    static void build(long[] arr)  
	    {            
	        // insert leaf nodes in tree 
	        for (int i = 0; i < n; i++)  
	            {tree[n + i] = arr[i]; ind[n+i]=i;}
	          
	        // build the tree by calculating 
	        // parents 
	        for (int i = n - 1; i > 0; i--)  
	        { if(tree[2*i] >= tree[2*i+1])
	            {tree[i] = tree[2*i]; ind[i] = ind[2*i];} 
	          else
	            {tree[i] = tree[2*i+1]; ind[i] = ind[2*i+1];} 
	        }
	    } 
	      
	    // function to update a tree node 
	    static void updateTreeNode(int p, long value)  
	    {  	          
	        // set value at position p 
	        tree[p+=n] = value;  
	          
	        // move upward and update parents 
	        for (int i = p/2; i > 0; i/=2) 
	          { if(tree[2*i] >= tree[2*i+1])
                  {tree[i] = tree[2*i]; ind[i] = ind[2*i];} 
                else
                  {tree[i] = tree[2*i+1]; ind[i] = ind[2*i+1];} 
              }
	    } 
	      
	    // function to get sum on 
	    // interval [l, r]
	    
	    static long query(int l, int r)  
	    {  
	        long res = 0; 
	          
	        // loop to find the sum/max/minimum/x-or etc  in the range 
	        
	        for (l+= n, r+= n; l <= r; l/=2, r/=2) 
	        { 
	            if (l%2==1) //if left end is a right child (odd index), we cannot include the parent,(as it will include elements out of range)  
	            	//so we have to include this node itself and move to the next node in range
	                {
	            	    if(tree[l]>res)
	            	       { res = tree[l];} 
	            	    
	            	    l++;
	                }
	          
	            if (r%2==0) //if right end is a left child(even index), we cannot include the parent,(as it will include elements out of range) 
	            	//so we have to include this node itself and move to the previous node in range
	            {
            	    if(tree[r]>res)
            	       { res = tree[r]; } 
            	    r--;
                }
          
	        } 
	          
	        return res; 
	    }
	    
	    static int queryIndex(int l, int r)  
	    {
	    	long res = 0; 
	        int in = -1; 
	        // loop to find the sum/max/minimum/x-or etc  in the range 
	        
	        for (l+= n, r+= n; l <= r; l/=2, r/=2) 
	        { 
	            if (l%2==1) //if left end is a right child (odd index), we cannot include the parent,(as it will include elements out of range)  
	            	//so we have to include this node itself and move to the next node in range
	                {
	            	    if(tree[l]>res)
	            	       { res = tree[l]; in = ind[l]; } 
	            	    if(tree[l]==res)
	            	       { if(ind[l] < in)
	            	    	   in = ind[l]; } 
	            	    l++;
	                }
	          
	            if (r%2==0) //if right end is a left child(even index), we cannot include the parent,(as it will include elements out of range) 
	            	//so we have to include this node itself and move to the previous node in range
	            {
            	   if(tree[r]>res)
            	       { res = tree[r]; in = ind[r]; } 
            	   if(tree[r]==res)
         	       { if(ind[r] < in)
         	    	   in = ind[r]; } 
            	    r--;
                }
          
	        } 
	        
	        return in;
	    }
	
	}
	
	public static void fill ( int low, int  high, SegmentTrees st, int[] d, int dep)
	{
		if(low>high)
			return;
		
		int in = st.queryIndex(low,high);
		d[in] = dep;
		fill ( low, in-1, st, d, dep+1);
		fill ( in+1, high, st, d, dep+1);
		
	}
	
	public static int[] solution(long[] arr, int n)
			
	{
		int[] d = new int[n];
		SegmentTrees st = new SegmentTrees(n);
		st.build(arr);
		fill (0, n-1, st, d, 0);
		return  d;				
	}
        
private static PrintWriter out = new PrintWriter(System.out);

public static void main (String[] args)
{
	MyScanner s =  new MyScanner();
     
    int t = s.nextInt();
    
    for(int j = 0; j<t ; j++)
    {
    	int n = s.nextInt();
    	long[] arr = new long[n];
    	for(int i =0; i<n; i++)
    		arr[i] = s.nextLong();
    	
    	int[] d = solution(arr,n);
    	for(int i= 0; i<n; i++)
    		out.print(d[i]+" ");
    	out.println();
    }
    
    out.flush();
    out.close();
    
}
}