A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).

For a positive integer n, we call a permutation p of length n good if the following condition holds for every pair i and j (1 ≤ i ≤ j ≤ n) — 

  * (p_i  OR  p_{i+1}  OR  …  OR  p_{j-1}  OR  p_{j}) ≥ j-i+1, where OR denotes the [bitwise OR operation.](https://en.wikipedia.org/wiki/Bitwise_operation#OR)



In other words, a permutation p is good if for every subarray of p, the OR of all elements in it is not less than the number of elements in that subarray. 

Given a positive integer n, output any good permutation of length n. We can show that for the given constraints such a permutation always exists.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.

The first and only line of every test case contains a single integer n (1 ≤ n ≤ 100).

Output

For every test, output any good permutation of length n on a separate line. 

Example

Input


3
1
3
7


Output


1
3 1 2
4 3 5 2 7 1 6

Note

For n = 3, [3,1,2] is a good permutation. Some of the subarrays are listed below. 

  * 3 OR 1 = 3 ≥ 2 (i = 1,j = 2) 
  * 3 OR 1 OR 2 = 3 ≥ 3 (i = 1,j = 3) 
  * 1 OR 2 = 3 ≥ 2 (i = 2,j = 3) 
  * 1 ≥ 1 (i = 2,j = 2) 



Similarly, you can verify that [4,3,5,2,7,1,6] is also good.
#include <bits/stdc++.h>
using namespace std;
int vis[1000][1000];
long long gn;
bool comp(pair<int, int> k, pair<int, int> j) {
  if (j.second < k.second)
    return true;
  else
    return false;
}
long long po(long long a, long long b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  if (b & 1) return a * po(a * a, (b - 1) / 2);
  if (b & 1 ^ 1) return po(a * a, b / 2);
  return 1;
}
int vectoint(std::vector<int> v) {
  reverse(v.begin(), v.end());
  long long p = 0;
  for (int i = 0; i < v.size(); i++) {
    p += v[i] * po(2, i);
  }
  return p;
}
int dp[100][100];
int print(int x, int y) {
  if (x <= 1) {
    if (y > 2 || y <= 0)
      return 0;
    else
      return 1;
  }
  if (dp[x][y] != 0)
    return dp[x][y];
  else {
    return dp[x][y] = print(x - 1, y) + print(x - 1, y - 1);
  }
}
void dele(stack<int>& s, int k) {
  if (k == 1) {
    s.pop();
    return;
  }
  int x = s.top();
  s.pop();
  dele(s, k - 1);
  s.push(x);
  return;
}
void sort0() {}
void finddup() {
  int arr[] = {1, 2, 3, 4, 5, 4};
  int tor = arr[0];
  int hare = arr[0];
  do {
    tor = arr[tor];
    hare = arr[arr[hare]];
  } while (hare != tor);
  hare = arr[0];
  while (hare != tor) {
    hare = arr[hare];
    tor = arr[tor];
  }
  cout << hare << endl;
  return;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    std::vector<int> v;
    for (int i = 1; i <= n; i++) cout << i << " ";
    cout << endl;
  }
  return 0;
}
