Problem: There are n cats in a line, labeled from 1 to n, with the i-th cat at position i. They are bored of gyrating in the same spot all day, so they want to reorder themselves such that no cat is in the same place as before. They are also lazy, so they want to minimize the total distance they move. Help them decide what cat should be at each location after the reordering.

For example, if there are 3 cats, this is a valid reordering: [3, 1, 2]. No cat is in its original position. The total distance the cats move is 1 + 1 + 2 = 4 as cat 1 moves one place to the right, cat 2 moves one place to the right, and cat 3 moves two places to the left.

Input

The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.

The first and only line of each test case contains one integer n (2 ≤ n ≤ 100) — the number of cats.

It can be proven that under the constraints of the problem, an answer always exist. 

Output

Output t answers, one for each test case. Each answer consists of n integers — a permutation with the minimum total distance. If there are multiple answers, print any.

Example

Input


2
2
3


Output


2 1 
3 1 2 

Note

For the first test case, there is only one possible permutation that satisfies the conditions: [2, 1].

The second test case was described in the statement. Another possible answer is [2, 3, 1].

Solution (in JAVA):

Solution:

import java.io.*;
import java.util.*;

public class Solution {

    /* -------------------------------------- Main Start ----------------------------------- */
    public static void main(String[] args) throws IOException {
        FastReader fs = new FastReader();
        int t = Integer.parseInt(fs.nextLine());
        while (t-- > 0) {
            int n = fs.nextInt();
            if (n % 2 == 0) {
                for (int i = 2; i <= n; i += 2) {
                    System.out.print(i + " " + (i - 1) + " ");
                }
                System.out.println("");
            } else {
                if (n == 3) {
                    System.out.println("2 3 1");
                } else {
                    for (int i = 2; i <= n - 3; i += 2) {
                        System.out.print(i + " " + (i - 1) + " ");
                    }
                    System.out.print((n - 1) + " " + (n) + " " + (n - 2));
                    System.out.println("");
                }
            }
        }

    }


    /* ---------------------------------------------------------------------------------------------------------------------------------------------------------------------- 
  
      StringBuilder sb = new StringBuilder();
      sb.append(x + "\n");
    
      Collections.sort(arr, (a, b) -> Integer.compare(a[0], b[0]))
      arr.toArray(new int[arr.size()][])
    
     */
    private static final int MOD_1 = 1000000000 + 7;

    static class FastReader {

        BufferedReader br;
        StringTokenizer st;

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

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

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

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

        double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

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

        /* --------------------------------------------------------------------------------------------------------------------------------------------------------------------- */
        void printArray(int[] arr) {
            System.out.println(Arrays.toString(arr));
        }

        void printArray(String[] arr) {
            System.out.println(Arrays.toString(arr));
        }

        void printArray(long[] arr) {
            System.out.println(Arrays.toString(arr));
        }

        void print(int data) {
            System.out.println(data);
        }

        void print(String data) {
            System.out.println(data);
        }

        void print(long data) {
            System.out.println(data);
        }

        int[] II(int n) throws IOException {
            int[] d = new int[n];
            String[] arr = nextLine().split(" ");
            for (int i = 0; i < n; i++) {
                d[i] = Integer.parseInt(arr[i]);
            }

            return d;
        }

        String[] IS(int n) throws IOException {
            return nextLine().split(" ");
        }

        long[] IL(int n) throws IOException {
            long[] d = new long[n];
            String[] arr = nextLine().split(" ");
            for (int i = 0; i < n; i++) {
                d[i] = Long.parseLong(arr[i]);
            }
            return d;
        }

        public long gcd(long a, long b) {
            if (a == 0) {
                return b;
            }
            return gcd(b % a, a);
        }

        public long power(long x, long y, long p) {
            long res = 1;
            x = x % p;
            if (x == 0) {
                return 0;
            }

            while (y > 0) {
                if ((y & 1) == 1) {
                    res = (res * x) % p;
                }
                y = y >> 1;
                x = (x * x) % p;
            }
            return res;
        }

        void sieveOfEratosthenes(boolean prime[], int size) {
            Arrays.fill(prime, true);
            prime[0] = prime[1] = false;
            prime[2] = true;

            for (int p = 2; p * p < size; p++) {
                if (prime[p] == true) {
                    for (int i = p * p; i < size; i += p) {
                        prime[i] = false;
                    }
                }
            }
        }

        public long fact(long n) {
            long ans = 1;
            for (int i = 2; i <= n; i++) {
                ans = (ans * i) % MOD_1;
            }
            return ans;
        }

        public long lcm(long a, long b) {
            return (a * b) / gcd(a, b);
        }

        public int log2(int N) {
            int result = (int) (Math.log(N) / Math.log(2));
            return result;
        }

        // find first element in the arr that is equal to the x. 
        // If x is not present in arr then it will return first element that is greater than x
        // It return arr.size() if all element are smaller than x
        public int lowerBound(int[] arr, int x) {
            int low = 0, high = arr.length, mid = -1;

            while (low < high) {
                mid = (low + high) / 2;

                if (arr[mid] >= x) {
                    high = mid;
                } else {
                    low = mid + 1;
                }
            }

            return low;
        }

        // find first element in the arr that is greater than x. 
        // It return arr.size() if all element are smaller than x
        public int upperBound(int[] arr, int x) {
            int low = 0, high = arr.length, mid = -1;

            while (low < high) {
                mid = (low + high) / 2;

                if (arr[mid] > x) {
                    high = mid;
                } else {
                    low = mid + 1;
                }
            }

            return low;
        }

    }

}