#include <bits/stdc++.h>
using namespace std;
const long long N = 1e5 + 5;
long long Ans = 1;
bool done[1000009];
vector<long long> primes;
void FindPrime(long long N) {
  done[1] = 1;
  for (int i = 2; i <= N; i++) {
    if (done[i]) continue;
    for (int j = i * 2; j <= N; j += i) {
      done[j] = 1;
    }
    primes.push_back(i);
  }
}
long long solve(long long Num) {
  for (int i = 0; i < primes.size(); i++) {
    int p = primes[i];
    if (p * p * p > Num) break;
    long long Count = 1;
    while (Num % p == 0) {
      Num /= p, Count++;
    }
    Ans *= Count;
  }
  long long sq = (long long)sqrt(Num);
  if (!done[Num])
    Ans *= 2;
  else if (sq * sq == Num && !done[sq])
    Ans *= 3;
  else if (Num != 1)
    Ans *= 4;
  return Ans;
}
long long fact[N], invfact[N];
long long pow(long long a, long long b, long long m) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans;
}
long long modinv(long long k) { return pow(k, 1000000007 - 2, 1000000007); }
long long nCr(long long x, long long y) {
  if (y > x) return 0;
  long long num = fact[x];
  num *= invfact[y];
  num %= 1000000007;
  num *= invfact[x - y];
  num %= 1000000007;
  return num;
}
void initinv() {
  fact[0] = 1;
  for (int i = 1; i <= N; i++) {
    fact[i] = (i * fact[i - 1]) % 1000000007;
  }
  invfact[N] = modinv(fact[N]);
  for (int i = N - 1; i >= 1; i--) {
    invfact[i] = ((i + 1) * invfact[i + 1]) % 1000000007;
  }
}
struct data {
  int lent;
  int x;
  int y;
};
struct cmp {
  bool operator()(data const& p1, data const& p2) {
    if (p1.lent != p2.lent) {
      return p1.lent < p2.lent;
    } else {
      return p1.x > p2.x;
    }
  }
};
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    cin >> n;
    int a[n + 1];
    data temp;
    temp.lent = n;
    temp.x = 1;
    temp.y = n;
    for (int i = 1; i <= n; i++) a[i] = -1;
    priority_queue<data, vector<data>, cmp> q;
    q.push(temp);
    int mx = 0;
    for (int i = 1; i <= n; i++) {
      data p = q.top();
      q.pop();
      if (p.x == p.y) {
        a[p.x] = i;
      } else if (p.x + 1 == p.y) {
        a[p.x] = i;
        temp.lent = 1;
        temp.x = p.y;
        temp.y = p.y;
        q.push(temp);
        mx = i;
      } else {
        mx = i;
        int mid = (p.x + p.y) / 2;
        a[mid] = i;
        int len = p.y - p.x + 1;
        if (len % 2 == 1) {
          int x = len / 2;
          int y = x - 1;
          temp.lent = x;
          temp.x = p.x;
          temp.y = mid - 1;
          q.push(temp);
          temp.lent = x;
          temp.x = mid + 1;
          temp.y = p.y;
          q.push(temp);
        } else {
          int x = len / 2;
          int y = x - 1;
          temp.lent = x;
          temp.x = mid + 1;
          temp.y = p.y;
          q.push(temp);
          temp.lent = y;
          temp.x = p.x;
          temp.y = mid - 1;
          q.push(temp);
        }
      }
    }
    for (int i = 1; i <= n; i++) cout << a[i] << " ";
    cout << "\n";
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
const long long N = 1e5 + 5;
long long Ans = 1;
bool done[1000009];
vector<long long> primes;
void FindPrime(long long N) {
  done[1] = 1;
  for (int i = 2; i <= N; i++) {
    if (done[i]) continue;
    for (int j = i * 2; j <= N; j += i) {
      done[j] = 1;
    }
    primes.push_back(i);
  }
}
long long solve(long long Num) {
  for (int i = 0; i < primes.size(); i++) {
    int p = primes[i];
    if (p * p * p > Num) break;
    long long Count = 1;
    while (Num % p == 0) {
      Num /= p, Count++;
    }
    Ans *= Count;
  }
  long long sq = (long long)sqrt(Num);
  if (!done[Num])
    Ans *= 2;
  else if (sq * sq == Num && !done[sq])
    Ans *= 3;
  else if (Num != 1)
    Ans *= 4;
  return Ans;
}
long long fact[N], invfact[N];
long long pow(long long a, long long b, long long m) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans;
}
long long modinv(long long k) { return pow(k, 1000000007 - 2, 1000000007); }
long long nCr(long long x, long long y) {
  if (y > x) return 0;
  long long num = fact[x];
  num *= invfact[y];
  num %= 1000000007;
  num *= invfact[x - y];
  num %= 1000000007;
  return num;
}
void initinv() {
  fact[0] = 1;
  for (int i = 1; i <= N; i++) {
    fact[i] = (i * fact[i - 1]) % 1000000007;
  }
  invfact[N] = modinv(fact[N]);
  for (int i = N - 1; i >= 1; i--) {
    invfact[i] = ((i + 1) * invfact[i + 1]) % 1000000007;
  }
}
struct data {
  int lent;
  int x;
  int y;
};
struct cmp {
  bool operator()(data const& p1, data const& p2) {
    if (p1.lent != p2.lent) {
      return p1.lent < p2.lent;
    } else {
      return p1.x > p2.x;
    }
  }
};
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    cin >> n;
    int a[n + 1];
    data temp;
    temp.lent = n;
    temp.x = 1;
    temp.y = n;
    for (int i = 1; i <= n; i++) a[i] = -1;
    priority_queue<data, vector<data>, cmp> q;
    q.push(temp);
    int mx = 0;
    for (int i = 1; i <= n; i++) {
      data p = q.top();
      q.pop();
      if (p.x == p.y) {
        a[p.x] = i;
      } else if (p.x + 1 == p.y) {
        a[p.x] = i;
        temp.lent = 1;
        temp.x = p.y;
        temp.y = p.y;
        q.push(temp);
        mx = i;
      } else {
        mx = i;
        int mid = (p.x + p.y) / 2;
        a[mid] = i;
        int len = p.y - p.x + 1;
        if (len % 2 == 1) {
          int x = len / 2;
          int y = x - 1;
          temp.lent = x;
          temp.x = p.x;
          temp.y = mid - 1;
          q.push(temp);
          temp.lent = x;
          temp.x = mid + 1;
          temp.y = p.y;
          q.push(temp);
        } else {
          int x = len / 2;
          int y = x - 1;
          temp.lent = x;
          temp.x = mid + 1;
          temp.y = p.y;
          q.push(temp);
          temp.lent = y;
          temp.x = p.x;
          temp.y = mid - 1;
          q.push(temp);
        }
      }
    }
    for (int i = 1; i <= n; i++) cout << a[i] << " ";
    cout << "\n";
  }
  return 0;
}

```