#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, i, j, k, l, m, b, t, sum, sum1, flag, max1, min1, tmp;
  cin >> t;
  while (t--) {
    cin >> n;
    long long a[n + 1];
    memset(a, 0, sizeof(a));
    priority_queue<pair<long long, pair<long long, long long>>> p;
    p.push({1, {1, n}});
    i = 1;
    while (!p.empty()) {
      pair<long long, pair<long long, long long>> pq;
      pq = p.top();
      l = pq.first;
      m = pq.second.first;
      b = pq.second.second;
      m = abs(m);
      b = abs(b);
      p.pop();
      a[(m + b) / 2] = i;
      i++;
      j = (m + b) / 2;
      m = j;
      if (m - 1 >= 1 and a[m - 1] == 0) {
        long long d = j;
        while (a[m - 1] == 0) {
          m--;
          if (m == 0) {
            m = 1;
            break;
          }
        }
        d--;
        p.push({d - m + 1, {-m, -d}});
      }
      if (j + 1 <= n and a[j + 1] == 0) {
        long long e = j;
        while (a[j + 1] == 0) {
          j++;
          if (j == n + 1) {
            j = n;
            break;
          }
        }
        e++;
        p.push({j - e + 1, {-e, -j}});
      }
    }
    for (i = 1; i <= n; i++) {
      cout << a[i] << " ";
    }
    cout << endl;
  }
}

### Prompt

Create a solution in Cpp for 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 mod = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, i, j, k, l, m, b, t, sum, sum1, flag, max1, min1, tmp;
  cin >> t;
  while (t--) {
    cin >> n;
    long long a[n + 1];
    memset(a, 0, sizeof(a));
    priority_queue<pair<long long, pair<long long, long long>>> p;
    p.push({1, {1, n}});
    i = 1;
    while (!p.empty()) {
      pair<long long, pair<long long, long long>> pq;
      pq = p.top();
      l = pq.first;
      m = pq.second.first;
      b = pq.second.second;
      m = abs(m);
      b = abs(b);
      p.pop();
      a[(m + b) / 2] = i;
      i++;
      j = (m + b) / 2;
      m = j;
      if (m - 1 >= 1 and a[m - 1] == 0) {
        long long d = j;
        while (a[m - 1] == 0) {
          m--;
          if (m == 0) {
            m = 1;
            break;
          }
        }
        d--;
        p.push({d - m + 1, {-m, -d}});
      }
      if (j + 1 <= n and a[j + 1] == 0) {
        long long e = j;
        while (a[j + 1] == 0) {
          j++;
          if (j == n + 1) {
            j = n;
            break;
          }
        }
        e++;
        p.push({j - e + 1, {-e, -j}});
      }
    }
    for (i = 1; i <= n; i++) {
      cout << a[i] << " ";
    }
    cout << endl;
  }
}

```