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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
bool sortBySecond(const pair<long long, long long> &a,
                  const pair<long long, long long> b) {
  return a.second > b.second;
}
int ncr[1001][1001];
bool vis_ncr[1001][1001];
long long NCR(int n, int r) {
  if (n < r) return 0;
  if (vis_ncr[n][r]) return ncr[n][r];
  if (r == 0) return 1;
  if (r == 1) return n;
  if (n == 0) return 0;
  ncr[n][r] = NCR(n - 1, r) + NCR(n - 1, r - 1);
  vis_ncr[n][r] = true;
  return ncr[n][r];
}
template <typename K, typename V>
void printMapMultipleValue(map<K, V> const &mp, string name) {
  cout << name << ":\n";
  for (auto const &pa : mp) {
    cout << pa.first << " : ";
    cout << "{";
    for (auto const &x : pa.second) cout << x << ',';
    cout << "}\n";
  }
}
template <typename K, typename V>
void printMap(map<K, V> const &mp, string name) {
  cout << name << ":" << endl;
  for (auto const &pa : mp)
    cout << "{" << pa.first << " : " << pa.second << "}" << endl;
}
bool isPrime[100006];
void _seive() {
  memset(isPrime, true, sizeof(isPrime));
  isPrime[1] = false;
  isPrime[0] = false;
  for (int i = 2; i * i <= 100005; i++) {
    for (int j = i * i; j <= 100005; j += i) {
      isPrime[j] = false;
    }
  }
}
long long expo(int a, int b) {
  if (b == 0)
    return 1;
  else if (b == 1)
    return a;
  else if (b % 2 == 0) {
    long long ret = expo(a, b / 2);
    ret = (ret % 1000000007 * ret % 1000000007) % 1000000007;
    return ret;
  } else {
    long long ret = expo(a, b / 2);
    ret = (ret % 1000000007 * ret % 1000000007) % 1000000007;
    ret = (ret % 1000000007 * a % 1000000007) % 1000000007;
    return ret;
  }
}
long long inverseMod(long long a, long long m) { return expo(a, m - 2) % m; }
long long fact[100005];
void factorial() {
  fact[0] = fact[1] = 1;
  for (int i = 2; i < 100005; i++)
    fact[i] = (fact[i - 1] % 1000000007 * i % 1000000007) % 1000000007;
}
long long GCD(long long a, long long b) {
  if (a == 0) return b;
  return GCD(b % a, a);
}
long long mult(long long a, long long b) {
  return (a % 1000000007 * b % 1000000007) % 1000000007;
}
long long add(long long a, long long b) {
  a += b;
  if (a >= 1000000007) return a - 1000000007;
  return a;
}
long long sub(long long a, long long b) {
  a -= b;
  if (a < 0) return a + 1000000007;
  return a;
}
struct cmp {
  bool operator()(const pair<int, pair<int, int>> &a,
                  const pair<int, pair<int, int>> &b) const {
    if (a.first != b.first)
      return a.first > b.first;
    else
      return a.second.first < b.second.first;
  }
};
void _run() {
  int n;
  cin >> n;
  set<pair<int, pair<int, int>>, cmp> s;
  int l = 0;
  int r = n - 1;
  vector<int> ans(n, 0);
  int cnt = 0;
  s.insert({n, {l, r}});
  while (s.size()) {
    cnt++;
    auto p = *s.begin();
    int x = p.second.first;
    int y = p.second.second;
    int mid = (x + y) / 2;
    ans[mid] = cnt;
    s.erase(s.begin());
    if (mid - 1 >= x) s.insert({mid - x, {x, mid - 1}});
    if (mid + 1 <= y) s.insert({y - mid, {mid + 1, y}});
  }
  for (auto it : ans) cout << it << ' ';
  cout << endl;
  ;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  int test = 1;
  cin >> test;
  while (test--) _run();
  return 0;
}