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.StringTokenizer;

public class Pretty {

    //Fast Input
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

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

    //Printing an array
    static void print(long[][] a, int r, int c) {
        PrintWriter pw = new PrintWriter(System.out);
        for (int i = 0; i < r; i++) {
            for (int j = 0; j < c; j++) {
                pw.print(a[i][j] + " ");
            }
            pw.println();
        }
        pw.flush();
    }

    //Used to generate random test cases
    //Must be changed according to input format
    static void testCaseGenerator(int t, int lowN, int highN, long lowA, long highA) throws FileNotFoundException {
        File input = new File("C:\\Users\\debar\\OneDrive\\Documents\\input.txt");
        PrintWriter pw = new PrintWriter(input);
        pw.println(t);
        for (int i = 0; i < t; i++) {
            int n = (int) (Math.random() * (highN - lowN)) + lowN;
            pw.println(n);
            for (int j = 0; j < n; j++) {
                long a = (long) (Math.random() * (highA - lowA)) + lowA;
                pw.print(a + " ");
            }
            pw.println();
        }
        pw.flush();
    }

    //Find factorial of numbers from 1 to n
    //Check the size of factorial array
    static long[] factorial(int n, long mod) {
        long[] factorial = new long[n];
        factorial[0] = 1;
        for (int i = 1; i <= n; i++) {
            factorial[i] = (factorial[i - 1] * i) % mod;
        }
        return factorial;
    }

    //Find factorial inverse from 1 to n
    //Check the size of inverse_factorial array
    static long[] factorialInverse(int n, long mod) {
        long[] inverse_factorial = modularInverse(n, mod);
        for (int i = 2; i <= n; i++) {
            inverse_factorial[i] = (inverse_factorial[i] * inverse_factorial[i - 1]) % mod;
        }
        return inverse_factorial;
    }

    //Inverse exist when two number are relatively prime
    //Based on Fermat little theorem (a^mod-1)%mod=1 when mod is prime
    //a * a^(mod - 2) % mod = 1
    //a^(mod - 2) % mod =a^-1
    //When mod != prime use Extended euclidean theorem to calculate x and y for equation a*x + mod*y = gcd(a,b)=1
    //(a*x + mod*y = gcd(a,b)=1) % mod = a*x %mod =1
    //https://www.geeksforgeeks.org/find-initial-integral-solution-of-linear-diophantine-equation-if-finite-solution-exists/?ref=rp
    //Used for only one number
    static long modularInverse(long a, long mod) {
        return binaryExponentiation(a, mod - 2, mod);
    }

    //Modular inverse for number from 1 to n - 1
    //Time complexity O(n) for n numbers
    //Proof https://cp-algorithms.com/algebra/module-inverse.html
    static long[] modularInverse(int n, long mod) {
        long[] inv = new long[n];
        inv[1] = 1;
        for (int i = 2; i < n; i++) {
            inv[i] = mod - (mod / i) * inv[(int) (mod % i)] % mod;
        }
        return inv;
    }

    //GCD calculation of a and b with complexity O(log(Max(a,b))
    static long gcd(long a, long b) {
        long r;
        while (b != 0) {
            r = a % b;
            a = b;
            b = r;
        }
        return a;
    }

    //Find prime number upto n with complexity O(n*log(log(n)))
    //Travel through the multiple of prime number and set it to false
    static boolean[] sieveOfEratosthenes(int n) {

        boolean[] prime = new boolean[n + 1];
        for (int i = 0; i <= n; i++)
            prime[i] = true;

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

        prime[0] = prime[1] = false;
        return prime;
    }

    //Find Prefix sum in O(n)
    static long[] PrefixSum(long[] arr, int n) {

        long[] prefixSum = new long[n];

        prefixSum[0] = arr[0];
        for (int i = 1; i < n; ++i)
            prefixSum[i] = prefixSum[i - 1] + arr[i];
        return prefixSum;
    }

    //Find a^n in O(log n)
    //If n=11 then n in binary is equal to 1011
    //a^n=a^(8+2+1) that is multiply with result when bit is set.
    //https://youtu.be/L-Wzglnm4dM
    static long binaryExponentiation(long a, long n, long mod) {
        long result = 1L;
        n = n % (mod - 1);
        //n = (mod - 1) * floor(n / (mod - 1)) + n % (mod - 1)
        //a^((mod - 1) * floor(n / (mod - 1))) % mod = 1 so a^n % mod =a^(n % (mod - 1)) % mod when mod is prime
        //Using fermat little theorem a^(p-1) % p = 1 when p is prime
        while (n > 0) {
            if ((n & 1) == 1) {
                result = (result * a) % mod;
            }
            a = (a * a) % mod;
            n /= 2;
        }
        return result;
    }

    //Matrix multiplication
    static long[][] matrixMultiply(long[][] a, long[][] b, int row, int col, int mid, long mod) {
        long[][] c = new long[row][col];
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                c[i][j] = 0;
                for (int k = 0; k < mid; k++) {
                    c[i][j] = (c[i][j] + (a[i][k] * b[k][j]) % mod) % mod;
                }
            }
        }
        return c;
    }

    //Matrix Exponentiation
    static long[][] matrixExponentiation(long[][] a, long n, int size, long mod) {
        if (n == 0) {
            long[][] temp = new long[size][size];
            for (int i = 0; i < size; i++) {
                temp[i][i] = 1;
            }
            return temp;//Returns identity matrix
        } else {
            long[][] temp = matrixExponentiation(a, n / 2, size, mod);
            if (n % 2 == 0) {
                return matrixMultiply(temp, temp, size, size, size, mod);
            } else {
                return matrixMultiply(matrixMultiply(temp, temp, size, size, size, mod), a, size, size, size, mod);
            }
        }
    }

    //Solution starts here
    public static void main(String[] args) {

        //Input output class reference
        FastReader sc = new FastReader();
        PrintWriter pw = new PrintWriter(System.out);

        //Number of test case
        int t = sc.nextInt();

        //For each test case
        for (int i = 0; i < t; i++) {
            int n=sc.nextInt();
            int a[]=new int[n];
            int temp;
            for(int j=0;j<n;j++)
            {
                a[j]=j+1;
            }
            for(int j=0;j<n-1;j+=2)
            {
                temp=a[j];
                a[j]=a[j+1];
                a[j+1]=temp;
            }
            if((n&1)==1)
            {
                temp=a[n-1];
                a[n-1]=a[n-2];
                a[n-2]=temp;
            }
            for (int j=0;j<n;j++)
            {
                pw.print(a[j]+" ");
            }
            pw.println();
        }
        pw.flush();
    }
    //Solution ends here
}