#include <bits/stdc++.h>
using namespace std;
const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
const int INF = 1000000000;
const long long LINF = 1000000000000000000;
const long long MOD = 998244353;
const double PI = acos(-1.0);
const double EPS = 1e-10;
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
void solve() {
  int n;
  cin >> n;
  set<int> st;
  st.insert(0);
  st.insert(n + 1);
  vector<int> ans(n + 1);
  priority_queue<pair<int, int>> pq;
  pq.emplace(n, 0);
  int cnt = 1;
  while (!pq.empty()) {
    auto [range, l] = pq.top();
    pq.pop();
    l *= -1;
    l++;
    auto itr = st.lower_bound(l);
    int r = *itr - 1;
    int x = 0;
    if ((r - l + 1) & 1)
      x = (r + l) / 2;
    else
      x = (r + l - 1) / 2;
    ans[x] = cnt;
    cnt++;
    if (x - l + 1 > 1) {
      pq.emplace(x - l, -l + 1);
    }
    if (*itr - x > 1) {
      pq.emplace(*itr - x - 1, -x);
    }
    st.insert(x);
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  cout << endl;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int T;
  cin >> T;
  for (int t = 0; t < T; t++) solve();
}

### Prompt

In Cpp, your task is to solve 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;
const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
const int INF = 1000000000;
const long long LINF = 1000000000000000000;
const long long MOD = 998244353;
const double PI = acos(-1.0);
const double EPS = 1e-10;
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
void solve() {
  int n;
  cin >> n;
  set<int> st;
  st.insert(0);
  st.insert(n + 1);
  vector<int> ans(n + 1);
  priority_queue<pair<int, int>> pq;
  pq.emplace(n, 0);
  int cnt = 1;
  while (!pq.empty()) {
    auto [range, l] = pq.top();
    pq.pop();
    l *= -1;
    l++;
    auto itr = st.lower_bound(l);
    int r = *itr - 1;
    int x = 0;
    if ((r - l + 1) & 1)
      x = (r + l) / 2;
    else
      x = (r + l - 1) / 2;
    ans[x] = cnt;
    cnt++;
    if (x - l + 1 > 1) {
      pq.emplace(x - l, -l + 1);
    }
    if (*itr - x > 1) {
      pq.emplace(*itr - x - 1, -x);
    }
    st.insert(x);
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  cout << endl;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int T;
  cin >> T;
  for (int t = 0; t < T; t++) solve();
}

```