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 
import java.io.*;
import java.util.*;
import java.lang.Math;

public class Main{
	static BufferedReader reader;
	static StringTokenizer tokenizer;
	static PrintWriter writer;

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

	static long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	static double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}

	static String nextToken() throws IOException {
		while (tokenizer == null || !tokenizer.hasMoreTokens()) {
			tokenizer = new StringTokenizer(reader.readLine());
		}
		return tokenizer.nextToken();
	}
	static String nexts() throws IOException {
	  	  tokenizer = new StringTokenizer(reader.readLine());
	   	 String s="";
	   	 while (tokenizer.hasMoreTokens()) {
			s+=tokenizer.nextElement()+" ";
		}
		return s;
	}
	public static int gcd(int x, int y){
		if (y == 0) return x; else return gcd(y, x % y);
	}

	public static boolean isPrime(int nn) 
	    { 
	         if(nn<=1){
		 return false; }
	        if(nn<=3){
		 return true; }
	      if (nn%2==0||nn%3==0){
		 return false; }
	        for (int i=5;i*i<=nn;i=i+6){ 
	            if(nn%i==0||nn%(i+2)==0){
	            	return false;  }
	            }
	         return true; 
	    } 

	public static void shuffle(int[] A) {
		for (int i = 0; i < A.length; i++) {
			int j = (int)(i * Math.random());
			int tmp = A[i];
			A[i] = A[j];
			A[j] = tmp;
		}
	}
	public static void shufflel(Long[] A) {
		for (int i = 0; i < A.length; i++) {
			int j = (int)(i * Math.random());
			long tmp = A[i];
			A[i] = A[j];
			A[j] = tmp;
		}
	}

	public static String reverse(String input)
 	   {
	        StringBuilder input2=new StringBuilder();
	        input2.append(input);
	        input2 = input2.reverse();
	        return input2.toString();
	    }


	 public static long power(int x, long n)
	{
	//    long mod = 1000000007;
		if (n == 0) {
			return 1;
		}

		long pow = power(x, n / 2);

		if ((n & 1) == 1) {
			return (x * pow * pow);
		}
		return (pow * pow);
	}
	public static long power(long x, long y, long p)     //x^y%p
	    { 
  	       long res = 1; 
	       x = x % p; 
  	       while (y > 0) { 
 	 	 if (y % 2 == 1) {
      	        	        res = (res * x) % p; 
		 }
  		 y = y >> 1; // y = y/2 
     	     	 x = (x * x) % p; 
     	         } 
  	         return res; 
    	} 

	 static long ncr(int n, int k)  
   	 { 
           		 long res = 1; 
  		   // Since C(n, k) = C(n, n-k)  
 		       if (k > n - k) { 
     		       k = n - k; 
      		  } 
  		  // Calculate value of [n*(n-1)*---*(n-k+1)] / [k*(k-1)*---*1]  
     		   for (int i = 0; i < k; ++i) { 
        		    res *= (n - i); 
         		   res /= (i + 1); 
     		   } 
  		  return res; 
 	   }
	public static int inv(int a, int m)     //Returns modulo inverse of a with respect to m using extended euclid algorithm
	    { 
	        int m0=m,t,q; 
	        int x0=0,x1=1; 
      	        if(m==1){
	                return 0; }
                 	         while(a>1) 
     	        {  
   	                q=a/m; 
    	                t=m; 
                                 m=a%m;a=t; 
    	                t=x0; 
     	                x0=x1-q*x0; 
      	                x1=t; } 
    	          if(x1<0){
 	   	 x1+=m0; }
   	           return x1; 
	    } 

	static int modInverse(int n, int p) 
  	{ 
    	          return (int)power((long)n, (long)(p - 2),(long)p); 
	 } 
	static int ncrmod(int n, int r, int p) 
   	 { 
	        if (r == 0)        // Base case 
	            return 1; 
  	        int[] fac = new int[n + 1];         // Fill factorial array so that we can find all factorial of r, n and n-r 
	        fac[0] = 1; 
	        for (int i = 1; i <= n; i++) 
                                     fac[i] = fac[i - 1] * i % p; 
   	         return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) % p;     //IMPORTANT
    	}	 

	static int lower(int a[], int x)   //Returns rightest index of x in sorted arr else -1
  	  {                                                //If not present returns index of just smaller element
  	            int n=a.length;
    	        int l=0;
                int r=n-1;
                int ans=-1;
                while(l<=r){
                    int m=(r-l)/2+l;
                    if(a[m]<=x){
                        ans=m;
                        l=m+1;
                    }
                    else{
                        r=m-1;
                    }
                }
                 return ans;  
                } 
	static int upper(int a[], int x)     //Returns leftest index of x in sorted arr else n
 	   {                                                  //If not present returns index of just greater element
 	            int n=a.length;
   	            int l=0;
                int r=n-1;
                int ans=n;
                while(l<=r){
                    int m=(r-l)/2+l;
                    if(a[m]>=x){
                        ans=m;
                        r=m-1;
                    }
                    else{
                         l=m+1;
                    }
                }
                 return ans; 
                 } 

	public static long stringtoint(String s){ // Convert bit string to number
		long a = 1;
		long ans = 0;
		StringBuilder sb = new StringBuilder();
		sb.append(s);  
		sb.reverse();
		s=sb.toString();
		for(int i=0;i<s.length();i++){
			ans = ans + a*(s.charAt(i)-'0');
			a = a*2;
		}
		return ans;
	}
 
	String inttostring(long a,long m){        // a is number and m is length of string required
		String res = "";                     // Convert a number in bit string representation
		for(int i=0;i<(int)m;i++){
			if((a&(1<<i))==1){
				res+='1';
			}else
				res+='0';
		}
		StringBuilder sb = new StringBuilder();
		sb.append(res);  
		sb.reverse();
		res=sb.toString();
		return res;
    }
	
	 static class R implements Comparable<R>{
	        int x, y;
	        public  R(int x, int y) {           //a[i]=new R(x,y);
	            this.x = x;
	            this.y = y;
	        }
	        public int compareTo(R o) {
            		return x-o.x;     //Increasing order(Which is usually required)
        		}
	    }   
/*   FUNCTION TO SORT HASHMAP ACCORDING TO VALUE
//USE IT AS FOLLOW	

//Make a new list
List<Map.Entry<Integer, Integer> > list = 
               new LinkedList<Map.Entry<Integer, Integer> >(h.entrySet()); 
  
        // Sort the list 
        Collections.sort(list, new Comparator<Map.Entry<Integer, Integer> >() { 
            public int compare(Map.Entry<Integer, Integer> o1,  
                               Map.Entry<Integer, Integer> o2) 
            { 
                return (o1.getValue()).compareTo(o2.getValue());     //This is used to sort string stored as VALUE, use below function here accordingly
            } 
        }); 
          
        // put data from sorted list to new hashmap  
        HashMap<String, Integer> temp = new LinkedHashMap<String, Integer>(); 
        for (Map.Entry<String, Integer> aa : list) { 
            temp.put(aa.getKey(), aa.getValue()); 
        } 
        return temp;   //Now temp is our new sorted hashmap
OR
	for(int i = 0;i<h.size();i++)  //accessing elements from the list made
		{
			int count = list.get(i).getValue();    //Value
			while(count >= 0)
			{
					int key=list.get(i).getKey();  //key
					count -- ;
			}
		}

////////////////////////////////////////////////////////////////////////////////////////
public int compare(Entry<Integer, Integer> o1, Entry<Integer, Integer> o2) 
	{
		if(o1.getValue() != o2.getValue())
			return o1.getValue() > o2.getValue()?1:-1;  //Sorts Value in increasing order
		else if(o1.getValue() == o2.getValue()){
			return o1.getKey() > o2.getKey() ? 1:-1;  //If value equal, then sorts key in increasing order
		}
		
		return Integer.MIN_VALUE;
	}
*/
//Check if palindrome string - System.out.print(A.equals(new StringBuilder(A).reverse().toString())?"Yes":"No");
	public static void main(String[] args) throws IOException {
		reader = new BufferedReader(new InputStreamReader(System.in));
		tokenizer = null;
		writer = new PrintWriter(System.out);
		solve();
		reader.close();
		writer.close();
	}
	
	private static void solve() throws IOException {
		int t = nextInt();
		while(t-->0){
			//String s= nextToken();
			//String[] s = str.split("\\s+");
			//ArrayList<Integer> ar=new ArrayList<Integer>();
			//HashSet<Integer> set=new HashSet<Integer>();
			//HashMap<Integer,String> h=new HashMap<Integer,String>();
			//R[] a1=new R[n];
			//char[] c=nextToken().toCharArray();

			int n = nextInt();
			//long n = nextLong();

			int[] a=new int[n];
			int[] ans=new int[n];
			//long[] a=new long[n];
			int j=-1;
			for(int i=0;i<n;i++){
				a[i]=nextInt();
				if(a[i]==n){
				    j=i;
				}
			}
			int m=0;
			for(int i=0;i<n;i++){
			    if(a[i]==n){
			        ans[i]=0;
			        m=1;
			        continue;
			    }
			    if(m==0){ 
			        int k=1;
			        int max=a[i];
			        for(int u=0;u<j;u++){
			            if(i==u){
			                max=a[i];
			                continue;
			            }
			            if(a[u]>max){
			                max=a[u];
			                k++;
			            }
			        }
			        ans[i]=k;
			    }
			    else{
			        int k=1;
			        int max=a[i];
			        for(int u=n-1;u>j;u--){
			            if(i==u){
			                max=a[i];
			                continue;
			            }
			            if(a[u]>max){
			                max=a[u];
			                k++;
			            }
			        }
			        ans[i]=k;
			    }
			}
			
			for(int i=0;i<n;i++){
			    writer.print(ans[i]+" ");
			}
            writer.println();
            
			
		 }
	}
}