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.BufferedReader;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.io.BufferedWriter;
import java.io.OutputStreamWriter;
import java.util.Comparator;
import java.util.PriorityQueue;

public class D {
    public static void main(String[] args) {
        FastReader reader = new FastReader();
        BufferedWriter log = new BufferedWriter(new OutputStreamWriter(System.out));
        int t;
        try {
            t = reader.nextInt();
            for (int i = 0; i < t; i++) {
                int n = reader.nextInt();

                if (n == 1) {
                    log.write("1\n");
                    log.flush();
                    continue;
                }

                int ans[] = new int[n + 1];

                PriorityQueue<Range> heap = new PriorityQueue<>(new Comparator<Range>() {
                    @Override
                    public int compare(Range o1, Range o2) {
                        int size1 = o1.r - o1.l + 1;
                        int size2 = o2.r - o2.l + 1;

                        if (size1 == size2) {
                            return Integer.compare(o1.l, o2.l);
                        }

                        return Integer.compare(size2, size1);
                    }
                });

                heap.add(new Range(1, n));

                int curr = 1;
                int one = 0;

                while (!heap.isEmpty()) {
                    Range range = heap.poll();
                    if (range.l == range.r) {
                        // Base case
                        ans[range.l] = curr;
                        curr++;
                        one--;
                        continue;
                    }

                    int mid = (range.l + range.r) / 2;
                    ans[mid] = curr;
                    Range range1 = new Range(range.l, mid - 1);
                    Range range2 = new Range(mid + 1, range.r);

                    if (mid - 1 >= range.l) {
                        heap.add(range1);
                        if (range1.l == range1.r) {
                            one++;
                        }
                    }
                    if (range.r >= mid + 1) {
                        heap.add(range2);
                        if (range2.l == range2.r) {
                            one++;
                        }
                    }

                    curr++;
                    if (one == heap.size()) {
                        break;
                    }
                }

                for (int j = 1; j <= n; j++) {
                    if (ans[j] == 0) {
                        ans[j] = curr;
                        curr++;
                    }
                }

                StringBuilder sb = new StringBuilder();
                for (int j = 1; j <= n; j++) {
                    sb.append(ans[j] + " ");
                }

                log.write(sb.toString() + "\n");
                log.flush();
            }
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

    static class Range {
        int l;
        int r;

        public Range(int l, int r) {
            this.l = l;
            this.r = r;
        }

        @Override
        public String toString() {
            return "Range{" +
                    "l=" + l +
                    ", r=" + r +
                    '}';
        }
    }
}
class FastReader {
    BufferedReader br;
    StringTokenizer st;

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

    public String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

    public int nextInt() {
        return Integer.parseInt(next());
    }

    public long nextLong() {
        return Long.parseLong(next());
    }

    public String nextLine() {
        String str = "";
        try {
            str = br.readLine();
        } catch (Exception e) {
            e.printStackTrace();
        }
        return str;
    }
}