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 CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;

#define ll          long long
#define int         ll
#define ld          long double
#define ff          first
#define ss          second
#define io          ios_base::sync_with_stdio(false); cin.tie(NULL);
#define yes         cout<<"YES"
#define no          cout<<"NO"
#define loop(i,a,n) for(int i=a;i<n;i++)
#define rloop(i,n,a)for(int i=n-1;i>=a;i--)
#define test        int tt=1; cin>>tt; while(tt--)
#define pb          push_back
#define pi          3.141592653589793238
#define all(v)      v.begin(),v.end()
#define vecl        vector<int>
#define pii         pair<int,int>
#define vpii        vector<pii>
const int MOD = 1e9 + 7;
int fast_expo(int x, int p) {
    if (p == 0) return 1;
    else if (p % 2 == 0) { int t = fast_expo(x, p / 2) % MOD; return (t * t) % MOD;}
    else return (x * (fast_expo(x, p - 1)) % MOD) % MOD;
}
int gcd(int a, int b) {
    if (a == 0 || b == 0) return a + b;
    else return gcd(b % a, a);
}

int lcm(int a, int b)
{
    return (a * b) / gcd(a, b);
}
vecl v;
void primeFactors(int n)
{
    // Print the number of 2s that divide n
    while (n % 2 == 0)
    {
        v.pb(2);
        n = n / 2;
    }

    // n must be odd at this point. So we can skip
    // one element (Note i = i +2)
    for (int i = 3; i <= sqrt(n); i = i + 2)
    {
        // While i divides n, print i and divide n
        while (n % i == 0)
        {
            v.pb(i);
            n = n / i;
        }
    }

    // This condition is to handle the case when n
    // is a prime number greater than 2
    if (n > 2)
        v.pb(n);
}
//int t[101][101];
void calc(vecl &a, int l, int r, vecl &b, int d)
{
    if (l >= r)
        return;

    int mx = 0, ans = 0;
    loop(i, l, r)
    {
        mx = max(mx, a[i]);
    }
    loop(i, l, r)
    {
        if (mx == a[i])
        {
            ans = i; break;
        }
    }
    b[ans] = d;
    calc(a, ans + 1, r, b, d + 1);
    calc(a, l, ans, b, d + 1);
}

void solve()
{
    int n;
    cin >> n;
    vecl a(n);
    loop(i, 0, n)
    {
        cin >> a[i];
    }
    //memset(t,-1, sizeof t);
    int l = 0, r = n;
    vecl ans(n + 1, 0);
    int d = 0;
    calc(a, l, r, ans, d);
    loop(i, 0, n)
    cout << ans[i] << " ";
}

signed main() {

    io
#ifndef ONLINE_JUDGE
    freopen("inputf.in", "r", stdin);
    freopen("outputf.out", "w", stdout);
#endif
    //calc();
    test
    {
        solve();
        cout << endl;
    }
    return 0;
}