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.ArrayList;
import java.util.Arrays;
import java.util.StringJoiner;
import java.util.StringTokenizer;

public class MainE {

    public static void main(String[] args) {
        var sc = new FastScanner(System.in);
        var pw = new PrintWriter(System.out);
        var T = sc.nextInt();
        for (int t = 0; t < T; t++) {
            var N = sc.nextInt();
            solve(N, pw);
        }
        pw.flush();
    }

    private static void solve(int N, PrintWriter pw) {
        // [1 2 3... N] を最初は1とN、あとは1づつ遠くになるように動かしていく

        long cost = 0;
        int[] A = new int[N];
        for (int i = 0; i < N; i++) {
            A[i] = i+1;
        }

        var op = new ArrayList<int[]>();
        op.add( new int[]{1, N} );
        cost += (long)(N-1) * (N-1);
        A[0] = N;
        A[N-1] = 1;

        // 前半(含む中間)をNに動かす
        for (int i = 1; i < (N+1)/2; i++) {
            op.add( new int[]{i+1, N} );
            int d = N-(i+1);
            cost += (long)d * d;

            int t = A[i];
            A[i] = A[N-1];
            A[N-1] = t;
        }

        // 後半を1に動かす
        for (int i = N-2; i >= (N+1)/2; i--) {
            op.add( new int[]{i+1, 1} );
            int d = i+1 - 1;
            cost += (long)d * d;

            int t = A[i];
            A[i] = A[0];
            A[0] = t;
        }

        pw.println(cost);
        writeSingleLine(A, pw);
        pw.println(op.size());
        for (int i = op.size()-1; i >= 0; i--) {
            var o = op.get(i);
            pw.println(o[0] + " " + o[1]);
        }
    }

    @SuppressWarnings("unused")
    static class FastScanner {
        private final BufferedReader reader;
        private StringTokenizer tokenizer;

        FastScanner(InputStream in) {
            reader = new BufferedReader(new InputStreamReader(in));
            tokenizer = null;
        }

        String next() {
            if (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        String nextLine() {
            if (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    return reader.readLine();
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken("\n");
        }

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

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

        int[] nextIntArray(int n) {
            var a = new int[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

        int[] nextIntArray(int n, int delta) {
            var a = new int[n];
            for (int i = 0; i < n; i++) a[i] = nextInt() + delta;
            return a;
        }

        long[] nextLongArray(int n) {
            var a = new long[n];
            for (int i = 0; i < n; i++) a[i] = nextLong();
            return a;
        }
    }

    static void writeLines(int[] as) {
        var pw = new PrintWriter(System.out);
        for (var a : as) pw.println(a);
        pw.flush();
    }

    static void writeLines(long[] as) {
        var pw = new PrintWriter(System.out);
        for (var a : as) pw.println(a);
        pw.flush();
    }

    static void writeSingleLine(int[] as, PrintWriter pw) {
        for (var i = 0; i < as.length; i++) {
            if (i != 0) pw.print(" ");
            pw.print(as[i]);
        }
        pw.println();
    }

    static void debug(Object... args) {
        var j = new StringJoiner(" ");
        for (var arg : args) {
            if (arg == null) j.add("null");
            else if (arg instanceof int[]) j.add(Arrays.toString((int[]) arg));
            else if (arg instanceof long[]) j.add(Arrays.toString((long[]) arg));
            else if (arg instanceof double[]) j.add(Arrays.toString((double[]) arg));
            else if (arg instanceof Object[]) j.add(Arrays.toString((Object[]) arg));
            else j.add(arg.toString());
        }
        System.err.println(j.toString());
    }
}