Problem: You are given an array a of length n consisting of zeros. You perform n actions with this array: during the i-th action, the following sequence of operations appears:

  1. Choose the maximum by length subarray (continuous subsegment) consisting only of zeros, among all such segments choose the leftmost one; 
  2. Let this segment be [l; r]. If r-l+1 is odd (not divisible by 2) then assign (set) a[(l+r)/(2)] := i (where i is the number of the current action), otherwise (if r-l+1 is even) assign (set) a[(l+r-1)/(2)] := i. 



Consider the array a of length 5 (initially a=[0, 0, 0, 0, 0]). Then it changes as follows:

  1. Firstly, we choose the segment [1; 5] and assign a[3] := 1, so a becomes [0, 0, 1, 0, 0]; 
  2. then we choose the segment [1; 2] and assign a[1] := 2, so a becomes [2, 0, 1, 0, 0]; 
  3. then we choose the segment [4; 5] and assign a[4] := 3, so a becomes [2, 0, 1, 3, 0]; 
  4. then we choose the segment [2; 2] and assign a[2] := 4, so a becomes [2, 4, 1, 3, 0]; 
  5. and at last we choose the segment [5; 5] and assign a[5] := 5, so a becomes [2, 4, 1, 3, 5]. 



Your task is to find the array a of length n after performing all n actions. Note that the answer exists and unique.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.

The only line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a.

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).

Output

For each test case, print the answer — the array a of length n after performing n actions described in the problem statement. Note that the answer exists and unique.

Example

Input


6
1
2
3
4
5
6


Output


1 
1 2 
2 1 3 
3 1 2 4 
2 4 1 3 5 
3 4 1 5 2 6 

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.util.stream.Stream;

public class Solution {

    private static final String NO = "NO";
    private static final String YES = "YES";
    public static BufferedWriter bw;
    public static FR sc;
    private static final int MOD = 1000000007;

    public static void main(String[] args) throws IOException {
        sc = new FR();
        bw = new BufferedWriter(new OutputStreamWriter(System.out));
        int t = 1;
        t = sc.i();
        while (t > 0) {
            t--;
            int n = sc.i();
            solve(n);
        }
        bw.close();

    }

    private static void solve(int n) throws IOException {
        int[] ans = new int[n];
        int diff = n;
        int x = 1;

        while (diff > 0) {
            int cur = 0;
            for (int i = 0; i < n; i++) {
                if (ans[i] == 0) {
                    cur++;
                } else {
                    if (cur == diff) {
                        if (diff % 2 == 1) {
                            ans[i - diff + (diff / 2)] = x++;
                        } else {
                            ans[i - diff - 1 + (diff / 2)] = x++;
                        }
                    }
                    cur = 0;
                }
            }
            if (cur == diff) {
                ans[n - 1 - (diff / 2)] = x++;
            }
            int max = 0;
            int tp = 0;
            for (int i = 0; i < n; i++) {
                if (ans[i] == 0) tp++;
                else
                    tp = 0;
                max = Math.max(tp, max);
            }
            diff = max;
        }
        pA(ans);
    }

    private static void pA(int[] ans) throws IOException {
        for (int an : ans) {
            bw.write(an + " ");
        }
        bw.write("\n");
    }


}


class FR {
    BufferedReader br;
    StringTokenizer st;

    public FR() {
        br = new BufferedReader(new InputStreamReader(System.in));
    }

    public String s() {
        if (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {

            }
        }
        return st.nextToken();
    }

    public int i() {
        return Integer.parseInt(s());
    }

    public Integer[] ia(int n) {
        Integer[] a = new Integer[n];
        for (int i = 0; i < n; i++) {
            a[i] = i();
        }
        return a;
    }

    public long[] la(int n) {
        long[] a = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = l();
        }
        return a;
    }

    public String[] sa(int n) {
        String[] a = new String[n];
        for (int i = 0; i < n; i++) {
            a[i] = s();
        }
        return a;
    }

    public long l() {
        return Long.parseLong(s());
    }
}