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:
	// Main Code at the Bottom
	import java.util.*;
	import java.io.*; 
	public class Main{
		//Fast IO class
	    static class FastReader {
	        BufferedReader br; 
	        StringTokenizer st; 
	        public FastReader() {
	        	boolean env=System.getProperty("ONLINE_JUDGE") != null;
	        	if(!env) {
	        		try {
						br=new BufferedReader(new FileReader("src\\input.txt"));
					} catch (FileNotFoundException e) {
						e.printStackTrace();
					}
	        	}
	        	else 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; 
	        } 
	    }     
	    static long MOD=(long)1e9+7;
	    //debug
	    static void debug(Object... o) {
	        System.out.println(Arrays.deepToString(o));
	    }
	    static FastReader sc=new FastReader();
	    static PrintWriter out=new PrintWriter(System.out);  
	    //Global variables and functions
	    //Main function(The main code starts from here)
	    public static void main (String[] args) throws java.lang.Exception {
	    	int test=1;
	    	test=sc.nextInt();
	    	while(test-->0) {
	    		int n=sc.nextInt();
	    		int ans[]=new int[n+1];
	    		int prev=1;
	    		long val=0L;
	    		for(int i=n/2+1;i<n;i++) {
	    			ans[prev]=i;
	    			prev=i;
	    			val+=(long)(i-1)*(long)(i-1);
	    		}
	    		ans[prev]=n;
	    		prev=n;
	    		for(int i=2;i<=n/2;i++) {
	    			ans[prev]=i;
	    			prev=i;
	    			val+=(long)(i-n)*(long)(i-n);
	    		}
	    		ans[prev]=1;
	    		val+=(long)(n-1)*(long)(n-1);
	    		out.println(val);
	    		for(int i=1;i<=n;i++) out.print(ans[i]+" ");
	    		out.println();
	    		out.println(n-1);
	    		for(int i=n/2+1;i<n;i++) {
	    			out.println(i+" "+1);
	    		}
	    		for(int i=2;i<=n/2;i++) out.println(i+" "+n);
	    		out.println(1+" "+n);
	    	}
	        out.flush();
	        out.close();
	    }
	}