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.math.BigInteger;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        for (int i = 0; i < t; i++) {
            int n = sc.nextInt();
            if (n == 2) {
                print2();
                continue;
            } else if (n == 3) {
                print3();
                continue;
            }
            BigInteger sum = BigInteger.ZERO;
            StringBuilder sbPerm = new StringBuilder();
            StringBuilder sbOperations = new StringBuilder();
            sbPerm.append(n / 2 + 1).append(' ');
            for (int j = 1; j < n - 1; j++) {
                int distance;
                if (j < n / 2) {
                    sbOperations.append(n / 2 + j).append(' ').append(1).append('\n');
                    sbPerm.append(j).append(' ');
                    distance = n - j - 1;
                } else {
                    if (j == n / 2 && n % 2 == 1) {
                        sbOperations.append(n / 2 + j).append(' ').append(1).append('\n');
                    } else {
                        sbOperations.append(n - j).append(' ').append(n).append('\n');
                    }
                    sbPerm.append(j + 2).append(' ');
                    distance = j;
                }
                sum = sum.add(BigInteger.valueOf(distance).pow(2));
            }
            sum = sum.add(BigInteger.valueOf(n - 1).pow(2));
            sbPerm.append(n / 2);
            sbOperations.append(1).append(' ').append(n);
            System.out.println(sum);
            System.out.println(sbPerm);
            System.out.println(n - 1);
            System.out.println(sbOperations);
//            sbPerm = new StringBuilder();
//            for (int j = 0; j < n - 1; j++) {
//                sbPerm.append(j + 1).append(' ').append(n).append('\n');
//            }
//            System.out.print(sbPerm);

        }
    }

    private static void print2() {
        System.out.println("1");
        System.out.println("2 1");
        System.out.println("1");
        System.out.println("2 1");
    }

    private static void print3() {
        System.out.println("5\n" +
                "2 3 1\n" +
                "2\n" +
                "1 3\n" +
                "3 2");
    }


}