Problem: Lunar rover finally reached planet X. After landing, he met an obstacle, that contains permutation p of length n. Scientists found out, that to overcome an obstacle, the robot should make p an identity permutation (make p_i = i for all i).

Unfortunately, scientists can't control the robot. Thus the only way to make p an identity permutation is applying the following operation to p multiple times: 

  * Select two indices i and j (i ≠ j), such that p_j = i and swap the values of p_i and p_j. It takes robot (j - i)^2 seconds to do this operation. 

Positions i and j are selected by the robot (scientists can't control it). He will apply this operation while p isn't an identity permutation. We can show that the robot will make no more than n operations regardless of the choice of i and j on each operation.

Scientists asked you to find out the maximum possible time it will take the robot to finish making p an identity permutation (i. e. worst-case scenario), so they can decide whether they should construct a new lunar rover or just rest and wait. They won't believe you without proof, so you should build an example of p and robot's operations that maximizes the answer.

For a better understanding of the statement, read the sample description.

Input

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

Each of next t lines contains the single integer n (2 ≤ n ≤ 10^5) – the length of p.

Note, that p is not given to you. You should find the maximum possible time over all permutations of length n.

It is guaranteed, that the total sum of n over all test cases doesn't exceed 10^5.

Output

For each test case in the first line, print how many seconds will the robot spend in the worst case.

In the next line, print the initial value of p that you used to construct an answer.

In the next line, print the number of operations m ≤ n that the robot makes in your example.

In the each of next m lines print two integers i and j — indices of positions that the robot will swap on this operation. Note that p_j = i must holds (at the time of operation).

Example

Input


3
2
3
3


Output


1
2 1
1
2 1
5
2 3 1
2
1 3
3 2
5
2 3 1
2
1 3
2 3

Note

For n = 2, p can be either [1, 2] or [2, 1]. In the first case p is already identity, otherwise robot will make it an identity permutation in 1 second regardless of choise i and j on the first operation.

For n = 3, p can be equals [2, 3, 1].

  * If robot will select i = 3, j = 2 on the first operation, p will become [2, 1, 3] in one second. Now robot can select only i = 1, j = 2 or i = 2, j = 1. In both cases, p will become identity in one more second (2 seconds in total).
  * If robot will select i = 1, j = 3 on the first operation, p will become [1, 3, 2] in four seconds. Regardless of choise of i and j on the second operation, p will become identity in five seconds. 



We can show, that for permutation of length 3 robot will always finish all operation in no more than 5 seconds.

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
public class EdA {
	static long[] mods = {1000000007, 998244353, 1000000009};
	static long mod = mods[0];
	public static MyScanner sc;
    public static PrintWriter out;
	public static void main(String[] omkar) throws Exception{
		// TODO Auto-generated method stub
 		sc = new MyScanner();
 		out = new PrintWriter(System.out);
 		int t = sc.nextInt();
 		for(int i = 0;i<t;i++){
//	 		String input1 = bf.readLine().trim();
//	 		String input2 = bf.readLine().trim();
//			COMPARING INTEGER OBJECTS U DO DOT EQUALS NOT ==
 			int n = sc.nextInt();
 			int[] perm = new int[n+1];
 			if (n%2 == 0){
 				for(int j = 2;j<=n/2;j++){
 					perm[j] = j-1;
 				}
 				for(int j = n/2+1;j<=n-1;j++){
 					perm[j] = j+1;
 				}
 				perm[1] = n/2+1;
 				perm[n] = n/2;
 				int[] original = new int[n+1];
 				for(int j =1 ;j<=n;j++){
 					original[j] = perm[j];
 				}
 				ArrayList<Integer> v1 = new ArrayList<>();
 				ArrayList<Integer> v2 = new ArrayList<>();
 				long sum = 0;
 				for(int j = 1;j<=n/2;j++){
 					if (j == n/2){
 						sum+=(long)(n-1)*(long)(n-1);
 						swap(1, n, perm);
 						v1.add(1);
 						v2.add(n);
 					}
 					else{
 						sum+=(long)(n-perm[n])*(long)(n-perm[n]);
 						v1.add(perm[n]);
 						v2.add(n);
 						swap(n, perm[n], perm);
 						sum+=(long)(perm[1]-1)*(long)(perm[1]-1);
 						v1.add(perm[1]);
 						v2.add(1);
 						swap(1, perm[1], perm);
 					}
 				}
 				out.println(sum);
 				for(int j =1 ;j<=n;j++){
 					out.print(original[j] + " ");
 				}
 				out.println();
 				out.println(v1.size());
 				for(int j = 0;j<v1.size();j++){
 					out.println(v1.get(j) + " " + v2.get(j));
 				}
 			}
 			else{
 				for(int j = 2;j<=n/2;j++){
 					perm[j] = j-1;
 				}
 				for(int j = n/2+2;j<=n-1;j++){
 					perm[j] = j+1;
 				}
 				perm[n/2+1] = n/2+1;
 				perm[1] = n/2+2;
 				perm[n] = n/2;
 				int temp1 = perm[1];
 				int temp2 = perm[n/2+1];
 				perm[1] = temp2;
 				perm[n/2+1] = temp1;
 				int[] original = new int[n+1];
 				for(int j =1 ;j<=n;j++){
 					original[j] = perm[j];
 				}
 				ArrayList<Integer> v1 = new ArrayList<>();
 				ArrayList<Integer> v2 = new ArrayList<>();
 				long sum = 0;
 				sum+=(long)(n/2)*(long)(n/2);
 				v1.add(n/2+1);
 				v2.add(1);
 				swap(1, n/2+1, perm);
 				for(int j = 1;j<=n/2;j++){
 					if (j == n/2){
 						sum+=(long)(n-1)*(long)(n-1);
 						swap(1, n, perm);
 						v1.add(1);
 						v2.add(n);
 					}
 					else{
 						sum+=(long)(n-perm[n])*(long)(n-perm[n]);
 						v1.add(perm[n]);
 						v2.add(n);
 						swap(n, perm[n], perm);
 						sum+=(long)(perm[1]-1)*(long)(perm[1]-1);
 						v1.add(perm[1]);
 						v2.add(1);
 						swap(1, perm[1], perm);
 					}
 				}
 				out.println(sum);
 				for(int j =1 ;j<=n;j++){
 					out.print(original[j] + " ");
 				}
 				out.println();
 				out.println(v1.size());
 				for(int j = 0;j<v1.size();j++){
 					out.println(v1.get(j) + " " + v2.get(j));
 				}
 				
 			}

//	 		for(int j = 0;j<array.length;j++){
//	 			out.print(array[j] + " ");
//	 		}
//	 		out.println();
 		}
	 		
 		out.close();
 	}
	public static void swap(int i, int j, int[] arr){
		int temp1 = arr[i];
		int temp2 = arr[j];
		arr[j] = temp1;
		arr[i] = temp2;
	}
	public static void sort(int[] array){
		ArrayList<Integer> copy = new ArrayList<>();
		for (int i : array)
			copy.add(i);
		Collections.sort(copy);
		for(int i = 0;i<array.length;i++)
			array[i] = copy.get(i);
	}
	static String[] readArrayString(int n){
		String[] array = new String[n];
		for(int j =0 ;j<n;j++)
			array[j] = sc.next();
		return array;
	}
	static int[] readArrayInt(int n){
    	int[] array = new int[n];
    	for(int j = 0;j<n;j++)
    		array[j] = sc.nextInt();
    	return array;
    }
	static int[] readArrayInt1(int n){
		int[] array = new int[n+1];
		for(int j = 1;j<=n;j++){
			array[j] = sc.nextInt();
		}
		return array;
	}
	static long[] readArrayLong(int n){
		long[] array = new long[n];
		for(int j =0 ;j<n;j++)
			array[j] = sc.nextLong();
		return array;
	}
	static double[] readArrayDouble(int n){
		double[] array = new double[n];
		for(int j =0 ;j<n;j++)
			array[j] = sc.nextDouble();
		return array;
	}
	static int minIndex(int[] array){
		int minValue = Integer.MAX_VALUE;
		int minIndex = -1;
		for(int j = 0;j<array.length;j++){
			if (array[j] < minValue){
				minValue = array[j];
				minIndex = j;
			}
		}
		return minIndex;
	}
	static int minIndex(long[] array){
		long minValue = Long.MAX_VALUE;
		int minIndex = -1;
		for(int j = 0;j<array.length;j++){
			if (array[j] < minValue){
				minValue = array[j];
				minIndex = j;
			}
		}
		return minIndex;
	}
	static int minIndex(double[] array){
		double minValue = Double.MAX_VALUE;
		int minIndex = -1;
		for(int j = 0;j<array.length;j++){
			if (array[j] < minValue){
				minValue = array[j];
				minIndex = j;
			}
		}
		return minIndex;
	}
	static long power(long x, long y){
		if (y == 0)
			return 1;
		if (y%2 == 1)
			return (x*power(x, y-1))%mod;
		return power((x*x)%mod, y/2)%mod;
	}
	static void verdict(boolean a){
        out.println(a ? "YES" : "NO");
    }
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;
        public MyScanner() {
            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;
        }
        
    }	
}

//StringJoiner sj = new StringJoiner(" "); 
//sj.add(strings)
//sj.toString() gives string of those stuff w spaces or whatever that sequence is