#include <bits/stdc++.h>
using namespace std;
int read() {
  int ans = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ans = ans * 10 + c - '0';
    c = getchar();
  }
  return ans * f;
}
const int N = 2e5 + 5;
int t, n, a[N], cnt;
struct Node {
  int l, r;
  bool operator<(const Node &x) const {
    if ((x.r - x.l + 1) == (r - l + 1)) return x.l < l;
    return (x.r - x.l + 1) > (r - l + 1);
  }
};
inline void work(int l, int r);
int main() {
  t = read();
  while (t--) {
    n = read();
    work(1, n);
    for (int i = 1; i <= n; ++i) printf("%d ", a[i]);
    printf("\n");
  }
  return 0;
}
inline void work(int l, int r) {
  cnt = 0;
  priority_queue<Node> qu;
  qu.push((Node){l, r});
  while (!qu.empty()) {
    l = qu.top().l;
    r = qu.top().r;
    qu.pop();
    if ((r - l + 1) % 2 == 1) {
      int mid = (l + r) >> 1;
      a[mid] = ++cnt;
      if (l <= mid - 1) qu.push((Node){l, mid - 1});
      if (mid + 1 <= r) qu.push((Node){mid + 1, r});
    } else {
      int mid = (l + r - 1) >> 1;
      a[mid] = ++cnt;
      if (l <= mid - 1) qu.push((Node){l, mid - 1});
      if (mid + 1 <= r) qu.push((Node){mid + 1, r});
    }
  }
}

### 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;
int read() {
  int ans = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ans = ans * 10 + c - '0';
    c = getchar();
  }
  return ans * f;
}
const int N = 2e5 + 5;
int t, n, a[N], cnt;
struct Node {
  int l, r;
  bool operator<(const Node &x) const {
    if ((x.r - x.l + 1) == (r - l + 1)) return x.l < l;
    return (x.r - x.l + 1) > (r - l + 1);
  }
};
inline void work(int l, int r);
int main() {
  t = read();
  while (t--) {
    n = read();
    work(1, n);
    for (int i = 1; i <= n; ++i) printf("%d ", a[i]);
    printf("\n");
  }
  return 0;
}
inline void work(int l, int r) {
  cnt = 0;
  priority_queue<Node> qu;
  qu.push((Node){l, r});
  while (!qu.empty()) {
    l = qu.top().l;
    r = qu.top().r;
    qu.pop();
    if ((r - l + 1) % 2 == 1) {
      int mid = (l + r) >> 1;
      a[mid] = ++cnt;
      if (l <= mid - 1) qu.push((Node){l, mid - 1});
      if (mid + 1 <= r) qu.push((Node){mid + 1, r});
    } else {
      int mid = (l + r - 1) >> 1;
      a[mid] = ++cnt;
      if (l <= mid - 1) qu.push((Node){l, mid - 1});
      if (mid + 1 <= r) qu.push((Node){mid + 1, r});
    }
  }
}

```