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.io.*;
import java.util.*;

public final class Solution {

    static PrintWriter out = new PrintWriter(System.out);
    static FastReader in = new FastReader();
    static long mod = (long) 1e9 + 7;
    static Pair[] moves = new Pair[]{new Pair(-1, 0), new Pair(1, 0), new Pair(0, -1), new Pair(0, 1)};

    public static void main(String[] args) {
        int t = i();

        out:
        while (t-- > 0) {
            int n = i();
            int[] arr = input(n);
            int[] depth = new int[n];
            fill(0, n - 1, 0, arr, depth);
            print(depth);
        }
        out.flush();
    }

    private static void fill(int l, int r, int d, int[] arr, int[] depth) {
        if (l > r) {
            return;
        }
        int maxIdx = l;
        int max = arr[l];
        for (int i = l + 1; i <= r; i++) {
            if (arr[i] > max) {
                max = arr[i];
                maxIdx = i;
            }
        }

        depth[maxIdx] = d;
        fill(l, maxIdx - 1, d + 1, arr, depth);
        fill(maxIdx + 1, r, d + 1, arr, depth);
    }

    static boolean isPS(double number) {
        double sqrt = Math.sqrt(number);
        return ((sqrt - Math.floor(sqrt)) == 0);
    }

    static int sd(long i) {
        int d = 0;
        while (i > 0) {
            d += i % 10;
            i = i / 10;
        }
        return d;
    }

    static int[] leastPrime;

    static void sieveLinear(int N) {
        int[] primes = new int[N];
        int idx = 0;
        leastPrime = new int[N + 1];
        for (int i = 2; i <= N; i++) {
            if (leastPrime[i] == 0) {
                primes[idx++] = i;
                leastPrime[i] = i;
            }
            int curLP = leastPrime[i];
            for (int j = 0; j < idx; j++) {
                int p = primes[j];
                if (p > curLP || (long) p * i > N) {
                    break;
                } else {
                    leastPrime[p * i] = p;
                }
            }
        }
    }


    static int lower(long A[], long x) {
        int l = -1, r = A.length;
        while (r - l > 1) {
            int m = (l + r) / 2;
            if (A[m] >= x) {
                r = m;
            } else {
                l = m;
            }
        }
        return r;
    }

    static int upper(long A[], long x) {
        int l = -1, r = A.length;
        while (r - l > 1) {
            int m = (l + r) / 2;
            if (A[m] <= x) {
                l = m;
            } else {
                r = m;
            }
        }
        return l;
    }

    static void swap(int A[], int a, int b) {
        int t = A[a];
        A[a] = A[b];
        A[b] = t;
    }

    static int lowerBound(int A[], int low, int high, int x) {
        if (low > high) {
            if (x >= A[high]) {
                return A[high];
            }
        }

        int mid = (low + high) / 2;

        if (A[mid] == x) {
            return A[mid];
        }

        if (mid > 0 && A[mid - 1] <= x && x < A[mid]) {
            return A[mid - 1];
        }

        if (x < A[mid]) {
            return lowerBound(A, low, mid - 1, x);
        }

        return lowerBound(A, mid + 1, high, x);
    }


    static long pow(long a, long b) {
        long pow = 1;
        long x = a;
        while (b != 0) {
            if ((b & 1) != 0) {
                pow = (pow * x) % mod;
            }
            x = (x * x) % mod;
            b /= 2;
        }
        return pow;
    }

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

    static void print(char A[]) {
        for (char c : A) {
            out.print(c);
        }
        out.println();
    }

    static void print(boolean A[]) {
        for (boolean c : A) {
            out.print(c + " ");
        }
        out.println();
    }

    static void print(int A[]) {
        for (int c : A) {
            out.print(c + " ");
        }
        out.println();
    }

    static void print(long A[]) {
        for (long i : A) {
            out.print(i + " ");
        }
        out.println();

    }

    static void print(List<Integer> A) {
        for (int a : A) {
            out.print(a + " ");
        }
    }

    static void printYes() {
        out.println("YES");
    }

    static void printNo() {
        out.println("NO");
    }

    static int i() {
        return in.nextInt();
    }

    static long l() {
        return in.nextLong();
    }

    static String s() {
        return in.nextLine();
    }

    static int[] input(int N) {
        int A[] = new int[N];
        for (int i = 0; i < N; i++) {
            A[i] = in.nextInt();
        }
        return A;
    }

    static long[] inputLong(int N) {
        long A[] = new long[N];
        for (int i = 0; i < A.length; i++) {
            A[i] = in.nextLong();
        }
        return A;
    }

    static int GCD(int a, int b) {
        if (b == 0) {
            return a;
        } else {
            return GCD(b, a % b);
        }
    }

    static long GCD(long a, long b) {
        if (b == 0) {
            return a;
        } else {
            return GCD(b, a % b);
        }
    }

}

class BIT {

    int n;
    int[] tree;

    public BIT(int n) {
        this.n = n;
        this.tree = new int[n + 1];
    }

    public static int lowbit(int x) {
        return x & (-x);
    }

    public void update(int x) {
        while (x <= n) {
            ++tree[x];
            x += lowbit(x);
        }
    }

    public int query(int x) {
        int ans = 0;
        while (x > 0) {
            ans += tree[x];
            x -= lowbit(x);
        }
        return ans;
    }

    public int query(int x, int y) {
        return query(y) - query(x - 1);
    }
}

class SegmentTree {

    long[] t;

    public SegmentTree(int n) {
        t = new long[n + n];
        Arrays.fill(t, Long.MIN_VALUE);
    }

    public long get(int i) {
        return t[i + t.length / 2];
    }

    public void add(int i, long value) {
        i += t.length / 2;
        t[i] = value;
        for (; i > 1; i >>= 1) {
            t[i >> 1] = Math.max(t[i], t[i ^ 1]);
        }
    }

    // max[a, b]
    public long max(int a, int b) {
        long res = Long.MIN_VALUE;
        for (a += t.length / 2, b += t.length / 2; a <= b; a = (a + 1) >> 1, b = (b - 1) >> 1) {
            if ((a & 1) != 0) {
                res = Math.max(res, t[a]);
            }
            if ((b & 1) == 0) {
                res = Math.max(res, t[b]);
            }
        }
        return res;
    }
}

class Pair {

    int i, j;

    Pair(int i, int j) {
        this.i = i;
        this.j = j;
    }
}

class FastReader {

    BufferedReader br;
    StringTokenizer st;

    public FastReader() {
        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;
    }


}