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:
//stan hu tao
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.util.*;
import java.io.*;
import java.math.*;

public class x1474E
{
    public static void main(String hi[]) throws Exception
    {
        BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
        int T = Integer.parseInt(infile.readLine());
        StringBuilder sb = new StringBuilder();
        while(T-->0)
        {
            int N = Integer.parseInt(infile.readLine());
            int[] arr = new int[N];
            int[] loc = new int[N];
            for(int i=0; i < N; i++)
                arr[i] = loc[i] = i;
            ArrayList<String> ls = new ArrayList<String>();
            ls.add("1 "+N);
            arr[0] = N-1;   arr[N-1] = 0;
            loc[0] = N-1;   loc[N-1] = 0;
            long cost = (long)(N-1)*(N-1);
            for(int i=1; i < N-1; i++)
            {
                int choice = 0;
                if(i < N-1-i)
                    choice = N-1;
                cost += (long)abs(choice-i)*abs(choice-i);
                ls.add((i+1)+" "+(choice+1));
                arr[i] = (arr[i]+arr[choice])-(arr[choice]=arr[i]);
                loc[i] = choice;    loc[arr[i]] = i;
            }
            sb.append(cost+"\n");
            for(int x: arr)
                sb.append((x+1)+" ");
            sb.append("\n");
            sb.append((N-1)+"\n");
            Collections.reverse(ls);
            for(String s: ls)
                sb.append(s+"\n");
        }
        System.out.print(sb);
    }
}