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// Divisible by i Problem Code: DIVBYISolvedSubmit

// You are given an integer N.

// Construct a permutation P of length N such that

// For all i (1≤i≤N−1), i divides abs(Pi+1−Pi).

// Recall that a permutation of length N is an array where every integer from 1 to N occurs exactly once.

// It can be proven that for the given constraints at least one such P always exists.

// Input Format

// The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follow.

// The only line of each test case contains an integer N - the length of the array to be constructed.

// Output Format

// For each test case, output a single line containing N space-separated integers P1,P2,…,PN, denoting the elements of the array P.

// If there exist multiple such arrays, print any.

// Constraints

// 1≤T≤5⋅104

// 2≤N≤105

// The sum of N over all test cases does not exceed 105.

// Sample Input 1

// 2

// 2

// 3

// Sample Output 1

// 1 2

// 2 1 3

// Explanation

// Test case 1: A possible array satisfying all the conditions is [1,2]:

// For i=1: abs(A2−A1)=abs(2−1)=1 is divisible by 1.

// Test case 2: A possible array satisfying all the conditions is [2,1,3]:

// For i=1: abs(A2−A1)=abs(1−2)=1 is divisible by 1.

#include <iostream>

#include<vector>

#include<algorithm>

using namespace std;

int main() {

int t;

cin>>t;

while(t--)

{

int n;

cin>>n;

if(n%2==1)

{

for(int i=n/2+1,j=n/2;j>0;j--,i++)

{

cout<<i<<" "<<j<<" ";

}

cout<<n<<endl;

}

else

{

for(int i=n/2,j=n/2+1;j<=n;j++,i--)

{

cout<<i<<" "<<j<<" ";

}

cout<<endl;

}

}

// your code goes here

return 0;

}