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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.nio.file.Paths;
import java.util.*;
public class Main extends PrintWriter {
    static BufferedReader s = new BufferedReader(new InputStreamReader(System.in));Main () { super(System.out); }public static void main(String[] args) throws IOException{ Main  d1=new Main   ();d1.main();d1.flush(); }

    void main() throws IOException {
        StringBuilder sb = new StringBuilder();

        int t = 1;
        t=i(s()[0]);
        while(t-->0) {
            String[] s1 = s();
            int n=i(s1[0]);int[] a=new int[n];arri(a,n);
            int max=0;int ind=0;
            for(int i=0;i<n;i++){
                 if(a[i]>max){
                     max=a[i];ind=i;
                 }
            }Stack<Integer> st=new Stack<>();
            int[] dp=new int[n];
            for(int i=0;i<n;i++){
                while(!st.isEmpty()&&a[st.peek()]<a[i]) st.pop();
                if(!st.isEmpty()) dp[i]=dp[st.peek()]+1;
                st.push(i);
            }st=new Stack<>();int[] dp1=new int[n];
            for(int i=n-1;i>=0;i--){
                while(!st.isEmpty()&&a[st.peek()]<a[i]) st.pop();
                if(!st.isEmpty()) dp1[i]+=dp1[st.peek()]+1;
                st.push(i);
            }

            for(int i=0;i<n;i++){
                sb.append((dp[i]+dp1[i])+" ");
            }sb.append("\n");
        }
        System.out.println(sb);
    }
    static int n1,n2,n3,n4;static long[] a1,a2,a3,a4;static HashMap<Integer,Integer>[] m1,m2,m3;
    static long[] dp1,dp2,dp3;
    public static long go(int i,int flag){
        if(flag==1)
        {
            if(dp1[i]!=-1) return dp1[i];
            long val=Integer.MAX_VALUE;
            for(int j=0;j<n2;j++){
                if(m1[i].containsKey(j)) continue;
                if(dp2[j]!=-1){
                    val=Math.min(val,a1[i]+dp2[j]);
                }else
                     val=Math.min(val,a1[i]+go(j,flag+1));
            }dp1[i]=val;
            return val;
        }else if(flag==2){
            if(dp2[i]!=-1) return dp2[i];
          //  System.out.println(i);
            long val=Integer.MAX_VALUE;
            for(int j=0;j<n3;j++){
                if(m2[i].containsKey(j)) continue;
                if(dp3[j]!=-1){
                    val=Math.min(val,a2[i]+dp3[j]);
                }else
                 val=Math.min(val,a2[i]+go(j,flag+1));
            }
            dp2[i]=val;
            return val;
        }else {
            if(dp3[i]!=-1) return dp3[i];
            long val=Integer.MAX_VALUE;
            for(int j=0;j<n4;j++){
                if(m3[i].containsKey(j)) continue;
                val=Math.min(val,a3[i]+a4[j]);
            }
            dp3[i]=val;
            return val;
        }

    }

    static String[] s() throws IOException { return s.readLine().trim().split("\\s+"); }
    static int i(String ss) {return Integer.parseInt(ss); }
    static long l(String ss) {return Long.parseLong(ss); }
    public void arr(long[] a,int n) throws IOException {String[] s2=s();for(int i=0;i<n;i++){ a[i]=l(s2[i]); }}
    public void arri(int[] a,int n) throws IOException {String[] s2=s();for(int i=0;i<n;i++){ a[i]=i(s2[i]); }}
}


