#include <bits/stdc++.h>
using namespace std;
long long int div_floor(long long int a, long long int b) {
  return a / b - (((a ^ b) < 0) and a % b);
}
long long int div_ceil(long long int a, long long int b) {
  return a / b + (((a ^ b) >= 0) and a % b);
}
long long int modpow(long long int a, long long int b,
                     long long int m = 1000000007) {
  a %= m;
  long long int res = 1;
  while (b) {
    if (b & 1) res = (res * a) % m;
    a = (a * a) % m;
    b >>= 1;
  }
  return res;
}
long long int bpow(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long int modinv(long long int a, long long int m = 1000000007) {
  return modpow(a, m - 2, m);
}
void graph(vector<vector<int> > &adj, int m) {
  int x, y;
  for (int i = 0; i < m; i++) {
    cin >> x >> y;
    adj[--x].push_back(--y);
    adj[y].push_back(x);
  }
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n, 0);
  priority_queue<pair<int, int> > q;
  q.push(make_pair(n, 0));
  int i = 1;
  while (!q.empty()) {
    pair<int, int> elem = q.top();
    q.pop();
    int s = -elem.second;
    int len = elem.first;
    int e = s + len - 1;
    int u = (s + e) / 2;
    a[u] = i++;
    if (u > s) q.push(make_pair(u - s, -s));
    if (e > u) q.push(make_pair(e - u, -(u + 1)));
  }
  for (auto elem : a) cout << elem << " ";
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) solve();
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

You are given an array a of length n consisting of zeros. You perform n actions with this array: during the i-th action, the following sequence of operations appears:

  1. Choose the maximum by length subarray (continuous subsegment) consisting only of zeros, among all such segments choose the leftmost one; 
  2. Let this segment be [l; r]. If r-l+1 is odd (not divisible by 2) then assign (set) a[(l+r)/(2)] := i (where i is the number of the current action), otherwise (if r-l+1 is even) assign (set) a[(l+r-1)/(2)] := i. 



Consider the array a of length 5 (initially a=[0, 0, 0, 0, 0]). Then it changes as follows:

  1. Firstly, we choose the segment [1; 5] and assign a[3] := 1, so a becomes [0, 0, 1, 0, 0]; 
  2. then we choose the segment [1; 2] and assign a[1] := 2, so a becomes [2, 0, 1, 0, 0]; 
  3. then we choose the segment [4; 5] and assign a[4] := 3, so a becomes [2, 0, 1, 3, 0]; 
  4. then we choose the segment [2; 2] and assign a[2] := 4, so a becomes [2, 4, 1, 3, 0]; 
  5. and at last we choose the segment [5; 5] and assign a[5] := 5, so a becomes [2, 4, 1, 3, 5]. 



Your task is to find the array a of length n after performing all n actions. Note that the answer exists and unique.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.

The only line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a.

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).

Output

For each test case, print the answer — the array a of length n after performing n actions described in the problem statement. Note that the answer exists and unique.

Example

Input


6
1
2
3
4
5
6


Output


1 
1 2 
2 1 3 
3 1 2 4 
2 4 1 3 5 
3 4 1 5 2 6 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long int div_floor(long long int a, long long int b) {
  return a / b - (((a ^ b) < 0) and a % b);
}
long long int div_ceil(long long int a, long long int b) {
  return a / b + (((a ^ b) >= 0) and a % b);
}
long long int modpow(long long int a, long long int b,
                     long long int m = 1000000007) {
  a %= m;
  long long int res = 1;
  while (b) {
    if (b & 1) res = (res * a) % m;
    a = (a * a) % m;
    b >>= 1;
  }
  return res;
}
long long int bpow(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long int modinv(long long int a, long long int m = 1000000007) {
  return modpow(a, m - 2, m);
}
void graph(vector<vector<int> > &adj, int m) {
  int x, y;
  for (int i = 0; i < m; i++) {
    cin >> x >> y;
    adj[--x].push_back(--y);
    adj[y].push_back(x);
  }
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n, 0);
  priority_queue<pair<int, int> > q;
  q.push(make_pair(n, 0));
  int i = 1;
  while (!q.empty()) {
    pair<int, int> elem = q.top();
    q.pop();
    int s = -elem.second;
    int len = elem.first;
    int e = s + len - 1;
    int u = (s + e) / 2;
    a[u] = i++;
    if (u > s) q.push(make_pair(u - s, -s));
    if (e > u) q.push(make_pair(e - u, -(u + 1)));
  }
  for (auto elem : a) cout << elem << " ";
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) solve();
}

```