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.
// package cp;
import java.io.*;
import java.math.*;
import java.util.*;
public class Cf_one {
	long gcd(long a,long b) {if(b==0)return a;else return gcd(b,a%b);}
	void swap(long a,long b) {long temp=a;a=b;b=temp;}
	StringBuilder sb=new StringBuilder();
	Integer[] ARR=new Integer[5];
	//Integer sort-TLE-Initialize object elements i.e a[0].
	long ceil(double n) {return (long)Math.ceil(1.0*n);}
	void divisor(long n,int start) {
		int cnt=0;for(int i=start;i<=Math.sqrt(n);i++) {
		if(n%i==0) {if(i==n/i)cnt++;else cnt+=2;}}}
	public static void main(String[] args) throws IOException {	
		PrintWriter out = new PrintWriter(System.out);
		Readers.init(System.in);
		Cf_one obj=new Cf_one();
		int t=Readers.nextInt();
		while(t-->0) {
			int n=Readers.nextInt();
			for (int i = 0; i <n; i++) {
				System.out.print((i+1)+" ");
			}
			System.out.println();
		}
		out.flush();
	}
	
}
class Readers {
 
    static BufferedReader reader;
 
    static StringTokenizer tokenizer;
 
 
 
    /** call this method to initialize reader for InputStream */
 
    static void init(InputStream input) {
 
        reader = new BufferedReader(
 
                     new InputStreamReader(input) );
 
        tokenizer = new StringTokenizer("");
 
    }
 
 
 
    /** get next word */
 
    static String next() throws IOException {
 
        while ( ! tokenizer.hasMoreTokens() ) {
 
            //TODO add check for eof if necessary
 
            tokenizer = new StringTokenizer(
 
                   reader.readLine() );
 
        }
 
        return tokenizer.nextToken();
 
    }
 
 
 
    static int nextInt() throws IOException {
 
        return Integer.parseInt( next() );
 
    }
 
	
 
    static double nextDouble() throws IOException {
 
        return Double.parseDouble( next() );
 
    }
    
    
    static long nextLong() throws IOException{
    	
		return Long.parseLong(next());
	}
 
}