#include <bits/stdc++.h>
using namespace std;
int ans[200005];
int n, val;
struct node {
  int l, r;
  int len;
  node(int lll, int rr, int llen) {
    l = lll;
    r = rr;
    len = llen;
  }
  friend bool operator<(const node &a, const node &b) {
    if (a.len == b.len) return a.l > b.l;
    return a.len < b.len;
  }
};
void bfs() {
  priority_queue<node> Q;
  node t = node(1, n, n);
  Q.push(t);
  while (!Q.empty()) {
    node T = Q.top();
    Q.pop();
    if (T.l > T.r) continue;
    int mid;
    int l = T.l, r = T.r;
    if ((r - l + 1) % 2 == 0)
      mid = (l + r - 1) / 2;
    else
      mid = (l + r) / 2;
    ans[mid] = val;
    val++;
    if (mid % 2 == 0) {
      if ((r - l + 1) % 2 == 0) {
        Q.push(node(mid + 1, r, r - mid));
        Q.push(node(l, mid - 1, mid - l));
      } else {
        Q.push(node(l, mid - 1, mid - l));
        Q.push(node(mid + 1, r, r - mid));
      }
    } else {
      if ((r - l + 1) % 2 == 0) {
        Q.push(node(mid + 1, r, r - mid));
        Q.push(node(l, mid - 1, mid - l));
      } else {
        Q.push(node(l, mid - 1, mid - l));
        Q.push(node(mid + 1, r, r - mid));
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cout.setf(ios::fixed), cout.precision(0);
  int t;
  cin >> t;
  while (t--) {
    memset(ans, 0, sizeof(ans));
    cin >> n;
    val = 1;
    if (n == 1)
      cout << 1 << "\n";
    else {
      bfs();
      cout << ans[1];
      for (int i = 2; i <= n; ++i) cout << " " << ans[i];
      cout << "\n";
    }
  }
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
int ans[200005];
int n, val;
struct node {
  int l, r;
  int len;
  node(int lll, int rr, int llen) {
    l = lll;
    r = rr;
    len = llen;
  }
  friend bool operator<(const node &a, const node &b) {
    if (a.len == b.len) return a.l > b.l;
    return a.len < b.len;
  }
};
void bfs() {
  priority_queue<node> Q;
  node t = node(1, n, n);
  Q.push(t);
  while (!Q.empty()) {
    node T = Q.top();
    Q.pop();
    if (T.l > T.r) continue;
    int mid;
    int l = T.l, r = T.r;
    if ((r - l + 1) % 2 == 0)
      mid = (l + r - 1) / 2;
    else
      mid = (l + r) / 2;
    ans[mid] = val;
    val++;
    if (mid % 2 == 0) {
      if ((r - l + 1) % 2 == 0) {
        Q.push(node(mid + 1, r, r - mid));
        Q.push(node(l, mid - 1, mid - l));
      } else {
        Q.push(node(l, mid - 1, mid - l));
        Q.push(node(mid + 1, r, r - mid));
      }
    } else {
      if ((r - l + 1) % 2 == 0) {
        Q.push(node(mid + 1, r, r - mid));
        Q.push(node(l, mid - 1, mid - l));
      } else {
        Q.push(node(l, mid - 1, mid - l));
        Q.push(node(mid + 1, r, r - mid));
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cout.setf(ios::fixed), cout.precision(0);
  int t;
  cin >> t;
  while (t--) {
    memset(ans, 0, sizeof(ans));
    cin >> n;
    val = 1;
    if (n == 1)
      cout << 1 << "\n";
    else {
      bfs();
      cout << ans[1];
      for (int i = 2; i <= n; ++i) cout << " " << ans[i];
      cout << "\n";
    }
  }
}

```