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;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
void printv(vector<long long int> v) {
  long long int sz = v.size();
  for (long long int i = 0; i < sz; i++) {
    if (i == sz - 1)
      cout << v[i] << "\n";
    else
      cout << v[i] << " ";
  }
}
void input(long long int a[], long long int sz) {
  for (long long int i = 0; i < sz; i++) cin >> a[i];
}
void print(long long int a[], long long int sz) {
  for (long long int i = 0; i < sz; i++) {
    if (i == sz - 1)
      cout << a[i] << "\n";
    else
      cout << a[i] << " ";
  }
}
long long int dx1[] = {1, 1, 0, -1, -1, -1, 0, 1};
long long int dy1[] = {0, 1, 1, 1, 0, -1, -1, -1};
long long int dx2[] = {1, 0, -1, 0};
long long int dy2[] = {0, 1, 0, -1};
long long int mul(long long int a, long long int b,
                  long long int p = 1000000007) {
  return ((a % p) * (b % p)) % p;
}
long long int add(long long int a, long long int b,
                  long long int p = 1000000007) {
  return (a % p + b % p) % p;
}
long long int power(long long int x, long long int y,
                    long long int p = 1000000007) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int modInverse(long long int n, long long int p = 1000000007) {
  return power(n, p - 2, p);
}
long long int fac[200005];
long long int inv[200005];
void fact(long long int p = 1000000007) {
  fac[0] = 1;
  for (long long int i = 1; i < 200005; i++) fac[i] = fac[i - 1] * i % p;
  for (long long int i = 0; i < 200005; i++) inv[i] = modInverse(fac[i]);
}
long long int ncrMod(long long int n, long long int r,
                     long long int p = 1000000007) {
  if (r == 0) return 1;
  long long int fac[n + 1];
  fac[0] = 1;
  for (long long int i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % p;
  return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
         p;
}
long long int convert(string num) {
  long long int sum = 0;
  for (int i = 0; i < num.size(); i++) {
    sum = sum * 10;
    sum += (num[i] - '0');
  }
  return sum;
}
bool ispeed2ime(long long int n) {
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
bool isPowerof2(long long int n) {
  while (n > 1) {
    if (n % 2 != 0) {
      return true;
    }
    n /= 2;
  }
  return false;
}
void cost(long long int start, long long int end,
          vector<pair<long long int, long long int>>& v) {
  if (start > end) {
    return;
  }
  long long int mid = (end + start) / 2;
  long long int gap = (end - start + 1);
  v.push_back({gap, mid});
  cost(start, mid - 1, v);
  cost(mid + 1, end, v);
}
bool cmp(pair<long long int, long long int> a,
         pair<long long int, long long int> b) {
  if (a.first == b.first) {
    return (a.second < b.second);
  } else {
    return (a.first > b.first);
  }
}
void solve() {
  long long int n;
  cin >> n;
  vector<pair<long long int, long long int>> v;
  cost(1, n, v);
  sort((v).begin(), (v).end(), cmp);
  long long int ans[n + 1];
  ans[0] = 0;
  long long int cnt = 1;
  for (int i = 0; i < v.size(); i++) {
    ans[v[i].second] = cnt;
    cnt++;
  }
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}