#include <bits/stdc++.h>
using namespace std;
long long max(long long a, long long b) { return (a > b) ? a : b; }
long long min(long long a, long long b) { return (a < b) ? a : b; }
long long gcd(long long a, long long b) {
  while (1) {
    if (a == 0) return b;
    b %= a;
    if (b == 0) return a;
    a %= b;
  }
}
int main() {
  long long n, m, i, j, t;
  t = 1;
  cin >> t;
  for (long long ii = 0; ii < t; ii++) {
    long long k;
    cin >> n;
    if (n == 1 || n == 2) {
      for (i = 0; i < n; i++) cout << i + 1 << " ";
      cout << endl;
      continue;
    }
    vector<vector<long long>> q[n + 1];
    vector<long long> b(2);
    b[0] = 1;
    b[1] = n;
    q[n].push_back(b);
    vector<long long> a(n + 1);
    j = 1;
    for (i = n; i > 0; i--) {
      sort(q[i].begin(), q[i].end());
      for (m = 0; m < q[i].size(); m++) {
        long long k = (q[i][m][1] + q[i][m][0]) / 2;
        a[k] = j;
        j++;
        long long g = q[i][m][1] - k;
        b[0] = k + 1;
        b[1] = q[i][m][1];
        q[g].push_back(b);
        g = k - q[i][m][0];
        b[1] = k - 1;
        b[0] = q[i][m][0];
        q[g].push_back(b);
      }
    }
    j = i;
    for (i = 1; i <= n; i++) {
      if (a[i] != 0)
        cout << a[i] << " ";
      else {
        cout << j << " ";
        j++;
      }
    }
    cout << endl;
  }
}

### Prompt

Generate 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 max(long long a, long long b) { return (a > b) ? a : b; }
long long min(long long a, long long b) { return (a < b) ? a : b; }
long long gcd(long long a, long long b) {
  while (1) {
    if (a == 0) return b;
    b %= a;
    if (b == 0) return a;
    a %= b;
  }
}
int main() {
  long long n, m, i, j, t;
  t = 1;
  cin >> t;
  for (long long ii = 0; ii < t; ii++) {
    long long k;
    cin >> n;
    if (n == 1 || n == 2) {
      for (i = 0; i < n; i++) cout << i + 1 << " ";
      cout << endl;
      continue;
    }
    vector<vector<long long>> q[n + 1];
    vector<long long> b(2);
    b[0] = 1;
    b[1] = n;
    q[n].push_back(b);
    vector<long long> a(n + 1);
    j = 1;
    for (i = n; i > 0; i--) {
      sort(q[i].begin(), q[i].end());
      for (m = 0; m < q[i].size(); m++) {
        long long k = (q[i][m][1] + q[i][m][0]) / 2;
        a[k] = j;
        j++;
        long long g = q[i][m][1] - k;
        b[0] = k + 1;
        b[1] = q[i][m][1];
        q[g].push_back(b);
        g = k - q[i][m][0];
        b[1] = k - 1;
        b[0] = q[i][m][0];
        q[g].push_back(b);
      }
    }
    j = i;
    for (i = 1; i <= n; i++) {
      if (a[i] != 0)
        cout << a[i] << " ";
      else {
        cout << j << " ";
        j++;
      }
    }
    cout << endl;
  }
}

```