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 PYTHON3):

Solution:
#pyrival orz
import os
import sys
import math
from io import BytesIO, IOBase


input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
    return(int(input()))
def inlt():
    return(list(map(int,input().split())))
def insr():
    s = input()
    return(list(s[:len(s) - 1]))
def invr():
    return(map(int,input().split()))

############ ---- Dijkstra with path ---- ############
def dijkstra(start, distance, path, n):
    # requires n == number of vertices in graph,
    # adj == adjacency list with weight of graph

    visited = [False for _ in range(n)]  # To keep track of vertices that are visited
    distance[start] = 0  # distance of start node from itself is 0

    for i in range(n):
        v = -1  # Initialize v == vertex from which its neighboring vertices' distance will be calculated
        for j in range(n):
            # If it has not been visited and has the lowest distance from start
            if not visited[v] and (v == -1 or distance[j] < distance[v]):
                v = j

        if distance[v] == math.inf:
            break

        visited[v] = True  # Mark as visited

        for edge in adj[v]:
            destination = edge[0]  # Neighbor of the vertex
            weight = edge[1]  # Its corresponding weight

            if distance[v] + weight < distance[destination]:  # If its distance is less than the stored distance
                distance[destination] = distance[v] + weight  # Update the distance
                path[destination] = v  # Update the path
        

def gcd(a, b):
    if b == 0:
        return a
    else:
        return gcd(b, a%b)

def lcm(a, b):
    return (a*b)//gcd(a, b)

def ncr(n, r):
    return math.factorial(n)//(math.factorial(n-r)*math.factorial(r))

def npr(n, r):
    return math.factorial(n)//math.factorial(n-r)

def seive(n):
    primes = [True]*(n+1)
    ans = []

    for i in range(2, n):
        if not primes[i]:
            continue

        j = 2*i
        while j <= n:
            primes[j] = False
            j += i

    for p in range(2, n+1):
        if primes[p]:
            ans += [p]

    return ans

def factors(n):
    factors = []

    x = 1
    while x*x <= n:

        if n % x == 0:
            if n // x == x:
                factors.append(x)
            else:
                factors.append(x)
                factors.append(n//x)

        x += 1

    return factors

# Functions: list of factors, seive of primes, gcd of two numbers,
# lcm of two numbers, npr, ncr

def main():
    try:
        for _ in range(inp()):
            n = inp()
            a = inlt()
            arr = [0]*n

            def solve(d, l, r):
                if l <= r:
                    mx = a.index(max(a[l:r+1]))
                    arr[mx] = d + 1
                else:
                    return

                solve(d+1, l, mx - 1)
                solve(d+1, mx+1, r)


            solve(-1, 0, n-1)

            print(*arr, sep = " ")

    except Exception as e:
        print(e)

        

# region fastio

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

# endregion

if __name__ == "__main__":
    main()