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.io.*;
import java.math.*;
import java.text.*;

public class Main {
    static PrintWriter out;
    static Reader in;
    public static void main(String[] args) throws IOException {
        input_output();
        //out = new PrintWriter(System.out);
        //in = new Reader(new FileInputStream("card.in"));
        Main solver = new Main();
        solver.solve();
        out.close(); 
        out.flush(); 
    }

    static int INF = (int)1e9+2;
    static int maxn = (int)1e6+5;
    static int mod = (int)1e9+7;
    static int n, m, q, t, k;
    static double pi = 3.141592653589;

    void solve() throws IOException{
        t = in.nextInt();

        while (t --> 0) {
            n = in.nextInt();

            long ans = 0;
            int[] p = new int[n+1];
            List<Integer> op = new ArrayList<>();

            int last = n;
            for (int i = n/2; i >= 2; i--) {
                p[last] = i;
                last = i;
                ans += (long)(n-i)*(n-i);
                op.add(i);
                op.add(n);
            }
            p[1] = n;
            ans += (long)(n-1)*(n-1);
            op.add(n);
            op.add(1);
            for (int i = n/2+1; i < n; i++) {
                p[last] = i;
                last = i;
                ans += (long)(i-1)*(i-1);
                op.add(i);
                op.add(1);
            }
            p[last] = 1;

            out.println(ans);
            for (int i = 1; i <= n; i++) out.print(p[i]+" ");
            out.println();
            out.println(n-1);
            for (int i = 0; i < op.size(); i+=2)
                out.println(op.get(i)+" "+op.get(i+1));
        }
    }

    //<>

    static class Reader {
 
        private InputStream mIs;
        private byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;
    
        public Reader() {
            this(System.in);
        }
    
        public Reader(InputStream is) {
            mIs = is;
        }
    
        public int read() {
            if (numChars == -1) {
                throw new InputMismatchException();
    
        }
            if (curChar >= numChars) {
                curChar = 0;
                try {
                    numChars = mIs.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (numChars <= 0) {
                    return -1;
                }
            }
            return buf[curChar++];
        }
    
        public String nextLine() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            } while (!isEndOfLine(c));
            return res.toString();
        }
    
        public String next() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            } while (!isSpaceChar(c));
            return res.toString();
        }
    
        double nextDouble()
        {
            return Double.parseDouble(next());
        }
    
        public long nextLong() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            long res = 0;
            do {
                if (c < '0' || c > '9') {
                    throw new InputMismatchException();
                }
                res *= 10;
                res += c - '0';
                c = read();
            } while (!isSpaceChar(c));
            return res * sgn;
        }
    
        public int nextInt() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            int res = 0;
            do {
                if (c < '0' || c > '9') {
                    throw new InputMismatchException();
                }
                res *= 10;
                res += c - '0';
                c = read();
            } while (!isSpaceChar(c));
            return res * sgn;
        }
    
        public boolean isSpaceChar(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }
    
        public boolean isEndOfLine(int c) {
            return c == '\n' || c == '\r' || c == -1;
        }
 
    }
    static void input_output() throws IOException {
        File f = new File("in.txt");
        if(f.exists() && !f.isDirectory()) { 
            in = new Reader(new FileInputStream("in.txt"));
        } else in = new Reader();
        f = new File("out.txt");
        if(f.exists() && !f.isDirectory()) {
            out = new PrintWriter(new File("out.txt"));
        } else out = new PrintWriter(System.out);
    }
}