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.util.Map.Entry;
import java.util.function.Function;

import javax.security.auth.kerberos.KerberosKey;

import org.omg.PortableServer.ServantActivator;

import java.math.*;
import java.time.temporal.ValueRange;
import java.awt.image.PixelGrabber;
import java.io.*;

public class Main {
	public static void main(String[] args) throws FileNotFoundException {
		InputReader in = new InputReader(System.in);
		// Scanner in = new Scanner(System.in);
		// Scanner in = new Scanner(new BufferedReader(new
		// InputStreamReader(System.in)));
		PrintWriter out = new PrintWriter(System.out);
		// InputReader in = new InputReader(new
		// File("ethan_traverses_a_tree.txt"));
		// PrintWriter out = new PrintWriter(new
		// File("ethan_traverses_a_tree-output.txt"));

		int pi = in.nextInt();
		for (int qi = 0; qi < pi; qi++) {
			int n = in.nextInt();
			if (n == 2) {
				out.printf("1\n");
				out.printf("2 1\n");
				out.printf("1\n");
				out.printf("1 2\n");
			} else if (n == 3) {
				out.printf("5\n");
				out.printf("2 3 1\n");
				out.printf("2\n");
				out.printf("1 3\n");
				out.printf("2 3\n");
			} else {
				long ans = (long) (n - 1) * (n - 1);
				for (int i = 2; i < n; i++) {
					long u = Math.max(i - 1, n - i);
					ans = ans + u * u;
				}

				int k = n / 2;
				if (n % 2 == 1) {
					k++;
				}
				int[] a = new int[n + 1];
				for (int i = 2; i <= k; i++) {
					a[i] = i + 1;
				}
				a[k] = 1;
				a[n] = 2;

				for (int i = k + 1; i < n; i++) {
					a[i] = i - 1;
				}
				a[k + 1] = n;
				a[1] = n - 1;

				out.printf("%d\n", ans);
				for (int i = 1; i < n; i++) {
					out.printf("%d ", a[i]);
				}
				out.printf("%d\n", a[n]);
				out.printf("%d\n", n - 1);
				for (int i = 2; i <= k; i++) {
					out.printf("%d %d\n", i, n);
				}
				for (int i = n - 1; i > k; i--) {
					out.printf("%d %d\n", i, 1);
				}
				out.printf("%d %d\n", 1, n);
			}
		}

		out.close();
	}

	static class InputReader {
		BufferedReader br;
		StringTokenizer st;

		public InputReader(File f) {
			try {
				br = new BufferedReader(new FileReader(f));
			} catch (FileNotFoundException e) {
				e.printStackTrace();
			}
		}

		public InputReader(InputStream in) {
			br = new BufferedReader(new InputStreamReader(in));
		}

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

		public boolean hasNext() {
			while (st == null || !st.hasMoreTokens()) {
				String s = null;
				try {
					s = br.readLine();
				} catch (IOException e) {
					e.printStackTrace();
				}
				if (s == null)
					return false;
				st = new StringTokenizer(s);
			}
			return true;
		}

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

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

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