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 
#include <bits/stdc++.h>
auto clk = clock();
const long double PI2 = acos(-1) / 2.0l;
const int dx[] = {0, 0, -1, 1}, dy[] = {-1, 1, 0, 0};
const int dx8[] = {-1, -1, -1, 0, 1, 1, 1, 0},
          dy8[] = {-1, 0, 1, 1, 1, 0, -1, -1};
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long MAXN = 100010;
long long Gcd(long long a, long long b) {
  if (a % b == 0) return b;
  return Gcd(b, a % b);
}
long long Max(long long a, long long b) {
  if (a > b) return a;
  return b;
}
long long Min(long long a, long long b) {
  if (a < b) return a;
  return b;
}
void kickstartFormat() {
  static int tno = 1;
  cout << "Case #" << tno++ << ": ";
}
void printvector(std::vector<long long> v) {
  for (int i = 0; i < v.size(); ++i) {
    cout << v[i] << " ";
  }
  cout << "" << endl;
}
void printarray(long long a[], long long n) {
  for (long long i = 0; i < n; ++i) {
    cout << a[i] << " ";
  }
  cout << "" << endl;
}
long long findlcm(vector<long long> arr, long long n) {
  long long ans = arr[0];
  for (int i = 1; i < n; i++) ans = (((arr[i] * ans)) / (Gcd(arr[i], ans)));
  return ans;
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long lowerbound(long long k, vector<long long> v) {
  long long l = 0;
  long long size = v.size();
  long long r = size - 1;
  long long mid;
  while (l <= r) {
    mid = (l + r) / 2;
    if (v[mid] < k)
      l = mid + 1;
    else {
      r = mid - 1;
    }
  }
  return l;
}
long long upperbound(long long k, vector<long long> v) {
  long long l = 0;
  long long size = v.size();
  long long r = size - 1;
  long long mid;
  while (l <= r) {
    mid = (l + r) / 2;
    if (v[mid] > k)
      r = mid - 1;
    else {
      l = mid + 1;
    }
  }
  return l;
}
long long spf[MAXN];
void sieve() {
  spf[1] = 1;
  for (int i = 2; i < MAXN; i++) spf[i] = i;
  for (int i = 4; i < MAXN; i += 2) spf[i] = 2;
  for (int i = 3; i * i < MAXN; i++) {
    if (spf[i] == i) {
      for (int j = i * i; j < MAXN; j += i)
        if (spf[j] == j) spf[j] = i;
    }
  }
}
long long n, m;
vector<long long> shortPath(vector<pair<long long, long long>> v[],
                            long long src, long long V = 0) {
  set<pair<long long, long long>> setds;
  long long sizee = V;
  long long IN = 1000000000000000000;
  vector<long long> dist(sizee, IN);
  vector<bool> vis(sizee, false);
  setds.insert(make_pair(0, src));
  dist[src] = 0;
  while (!setds.empty()) {
    pair<long long, long long> tmp = *(setds.begin());
    setds.erase(setds.begin());
    long long u = tmp.second;
    if (vis[u]) continue;
    vis[u] = true;
    vector<pair<long long, long long>>::iterator i;
    for (auto i = v[u].begin(); i != v[u].end(); ++i) {
      long long v = (*i).first;
      long long weight = (*i).second;
      if (dist[v] > dist[u] + weight) {
        if (dist[v] != 1000000000000000000)
          setds.erase(setds.find(make_pair(dist[v], v)));
        dist[v] = dist[u] + weight;
        setds.insert(make_pair(dist[v], v));
      }
    }
  }
  return dist;
}
long long root(long long a[], long long r) {
  if (a[r] == r) {
    return r;
  } else {
    long long x = root(a, a[r]);
    a[r] = x;
    return x;
  }
}
void unio(long long a[], long long size[], long long x, long long y) {
  x = root(a, x);
  y = root(a, y);
  if (x == y) {
    return;
  }
  if (size[x] >= size[y]) {
    a[y] = x;
    size[x] += size[y];
  } else {
    a[x] = y;
    size[y] += size[x];
  }
}
vector<long long> d(110);
vector<long long> ar(110);
void rec(long long i, long long j, long long l) {
  if (i < 0 or j >= n or i > j) {
    return;
  }
  long long ind = 0;
  long long mx = -1;
  for (long long z = i; z <= j; z++) {
    mx = max(ar[z], mx);
  }
  for (long long z = i; z <= j; z++) {
    if (mx == ar[z]) {
      ind = z;
      break;
    }
  }
  d[ind] = l;
  rec(i, ind - 1, l + 1);
  rec(ind + 1, j, l + 1);
  return;
}
void CV(long long txt) {
  cin >> n;
  for (int i = 0, _n = (n); i < _n; i++) {
    cin >> ar[i];
  }
  for (int i = 0, _n = (n); i < _n; i++) {
    cout << ar[i] << " ";
  }
  cout << endl;
  rec((long long)0, n - 1, (long long)0);
  for (int i = 0, _n = (n); i < _n; i++) {
    cout << d[i] << " ";
  }
  cout << endl;
  cerr << endl;
  for (int i = 0, _n = (n); i < _n; i++) {
    ar[i] = d[i] = 0;
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long x68 = 1;
  cin >> x68;
  for (long long vi = 1; vi <= x68; vi++) {
    CV(vi);
  }
  return 0;
}
