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.*;
import java.io.*;

public class A
{
   public static void main(String[] args) throws Exception
   {
      PrintWriter out = new PrintWriter(System.out);
      new A(new FastScanner(System.in), out);
      out.close();
   }

   public A(FastScanner in, PrintWriter out)
   {
			 int t = in.nextInt();
			 while(t-->0) 
			 {
			 		int n = in.nextInt();
					for (int i=0; i<n; i++) 
					{
							System.out.print(i+1+" ");
					} 
					System.out.println();		
			 }
   }
}

class FastScanner{
    private InputStream stream;
    private byte[] buf = new byte[1024];
    private int curChar;
    private int numChars;
    
    public FastScanner(InputStream stream)
    {
        this.stream = stream;
    }
    
    int read()
    {
        if (numChars == -1)
            throw new InputMismatchException();
        if (curChar >= numChars){
            curChar = 0;
            try{
                numChars = stream.read(buf);
            } catch (IOException e) {
                throw new InputMismatchException();
            }
            if (numChars <= 0)
                return -1;
        }
        return buf[curChar++];
    }
    
    boolean isSpaceChar(int c)
    {
        return c==' '||c=='\n'||c=='\r'||c=='\t'||c==-1;
    }
    
    boolean isEndline(int c)
    {
        return c=='\n'||c=='\r'||c==-1;
    }
    
    int nextInt()
    {
        return Integer.parseInt(next());
    }
    
    long nextLong()
    {
        return Long.parseLong(next());
    }
    
    double nextDouble()
    {
        return Double.parseDouble(next());
    }
    
    String next(){
        int c = read();
        while (isSpaceChar(c))
            c = read();
        StringBuilder res = new StringBuilder();
        do{
            res.appendCodePoint(c);
            c = read();
        }while(!isSpaceChar(c));
        return res.toString();
    }
    
    String nextLine(){
        int c = read();
        while (isEndline(c))
            c = read();
        StringBuilder res = new StringBuilder();
        do{
            res.appendCodePoint(c);
            c = read();
        }while(!isEndline(c));
        return res.toString();
    }
}

