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.io.*;
import java.util.*;

public class CF1474E extends PrintWriter {
	CF1474E() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1474E o = new CF1474E(); o.main(); o.flush();
	}

	int[] ii, jj; int cnt; long ans;
	void add(int i, int j) {
		ii[cnt] = i; jj[cnt] = j; cnt++;
		int d = Math.abs(j - i);
		ans += (long) d * d;
	}
	void main() {
		int t = sc.nextInt();
		while (t-- > 0) {
			int n = sc.nextInt();
			ii = new int[n];
			jj = new int[n];
			cnt = 0; ans = 0;
			int m = (n + 1) / 2, l = m, r = m;
			add(m, n);
			while (cnt < n - 1)
				if (n - l <= r - 1)
					add(--l, n);
				else
					add(++r, 1);
			println(ans);
			int[] pp = new int[n + 1];
			for (int i = 1; i <= n; i++)
				pp[i] = i;
			for (int h = cnt - 1; h >= 0; h--) {
				int i = ii[h], j = jj[h];
				int tmp = pp[i]; pp[i] = pp[j]; pp[j] = tmp;
			}
			for (int i = 1; i <= n; i++)
				print(pp[i] + " ");
			println();
			println(cnt);
			for (int h = 0; h < cnt; h++)
				println(ii[h] + " " + jj[h]);
		}
	}
}