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.
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Scanner;
import java.util.Set;
import java.util.stream.Collectors;

public class Solution {
  static int n;

  public static void main(String args[]) {
    Scanner in = new Scanner(System.in);
    int N = 1;
    N = in.nextInt();
    ext:
    for (int i = 0; i < N; i++) {
      n = 3;
      n = in.nextInt();
      List<int[]> ps = new ArrayList<>();
      ps.add(new int[n]);
      while (!ps.isEmpty()) {
        int[] p = ps.remove(0);
        loop:
        for (int k = 1; k <= n; k++) {
          int k2 = 0;
          for (; p[k2] != 0; k2++) {
            if (p[k2] == k) {
              continue loop;
            }
          }
          int[] newP = new int[n];
          System.arraycopy(p, 0, newP, 0, n);
          newP[k2] = k;
          if (k2 == n - 1) {
//            System.out.println(Arrays.toString(newP) );
//            ps.forEach(t -> System.out.println(Arrays.toString(t)));
            if (testGood(newP)) {
              continue ext;
            }

          } else {
            ps.add(0, newP);
          }
        }
      }

      //      ps.forEach(t -> System.out.println(Arrays.toString(t)));
    }
    //  System.out.println(3|2 );
    //  System.out.println(2|3);
  }

  public static boolean testGood(int[] p) {
    for (int I = 0; I < n; I++) {
      for (int J = I; J < n; J++) {
        int r = p[I];
        for (int k = I + 1; k <= J; k++) {
          r = r | p[k];
        }
        //        System.out.println(
        //            r + " " + (J - I + 1) + " " + Arrays.toString(p) + " " + I + " " + J);
        if (r < J - I + 1) {
          //          System.out.println(
          //              r + " " + (J - I + 1) + " " + Arrays.toString(p) + " " + I + " " + J);
          return false;
        }
      }
    }
    //    StringBuilder sb = new StringBuilder();
    //    for (int j : p) {
    //      sb.append(j + " ");
    //    }
    System.out.println(Arrays.stream(p).mapToObj(t -> t + "").collect(Collectors.joining(" ")));
    return true;
  }
}
