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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;
import java.util.stream.Stream;

public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        solve(in, out);
        out.close();
    }

    static String reverse(String s) {
        return (new StringBuilder(s)).reverse().toString();
    }

    static void sieveOfEratosthenes(int n, int factors[]) 
    { 
        factors[1]=1;
        for(int p = 2; p*p <=n; p++) 
        { 
            if(factors[p] == 0) 
            { 
                factors[p]=p;
                for(int i = p*p; i <= n; i += p) 
                    factors[i] = p; 
            } 
        } 
    }

    static void sort(int ar[]) {
        int n = ar.length;
        ArrayList<Integer> a = new ArrayList<>();
        for (int i = 0; i < n; i++)
            a.add(ar[i]);
        Collections.sort(a);
        for (int i = 0; i < n; i++)
            ar[i] = a.get(i);
    }

    static void sort1(long ar[]) {
        int n = ar.length;
        ArrayList<Long> a = new ArrayList<>();
        for (int i = 0; i < n; i++)
            a.add(ar[i]);
        Collections.sort(a);
        for (int i = 0; i < n; i++)
            ar[i] = a.get(i);
    }

    static long ncr(long n, long r, long mod) {
        if (r == 0)
            return 1;
        long val = ncr(n - 1, r - 1, mod);
        val = (n * val) % mod;
        val = (val * modInverse(r, mod)) % mod;
        return val;
    }

    public static void solve(InputReader sc, PrintWriter pw) {
        int i, j = 0;
        // int t = 1; 
        long sum[] = new long[100001];
        for(i=1;i<100001;i++){
            sum[i] = sum[i-1] + 1L*i*i;
        }
        int t = sc.nextInt();
        u: while (t-- > 0) {
            int n = sc.nextInt();
            if(n%2==0){
                pw.println((sum[n-1]-sum[n-n/2-1])*2-(1L*(n-1)*(n-1)));
                pw.print(n/2+1+" ");
                for(i=1;i<n/2;i++){
                    pw.print(i+" ");
                }
                for(i=n/2+2;i<=n;i++){
                    pw.print(i+" ");
                }
                pw.println(n/2);
                pw.println(n-1);
                for(i=0;i<n/2-1;i++)
                    pw.println((n/2+1+i)+" "+1);
                for(i=0;i<n/2;i++)
                    pw.println((n-n/2-i)+" "+n);
            }
            else{
                pw.println((sum[n-1]-sum[n-n/2-2])*2-(1L*(n-1)*(n-1))-(1L*(n-n/2-1)*(n-n/2-1)));
                int y = n/2+1;
                pw.print(y+1+" ");
                for(i=1;i<y;i++){
                    pw.print(i+" ");
                }
                for(i=y+2;i<=n;i++){
                    pw.print(i+" ");
                }
                pw.println(y);
                pw.println(n-1);
                for(i=y+1;i<n;i++)
                    pw.println(i+" "+1);
                for(i=n-y+1;i>=1;i--)
                    pw.println(i+" "+n);
            }
        }
    }

    static class Pair1 {
        long a;
        long b;
        
        Pair1(long a, long b) {
            this.a = a;
            this.b = b;
        }
    }

    static class Pair implements Comparable<Pair> {
        int a;
        int b;
        int i;
        
        Pair(int a, int b, int i) {
            this.a = a;
            this.b = b;
            this.i = i;
        }
 
        public int compareTo(Pair p) {
            if(a!=p.a)
                return a-p.a;
            return b-p.b;
        }
    }

    static boolean isPrime(long n) {
        if (n <= 1)
            return false;
        if (n <= 3)
            return true;
        if (n % 2 == 0 || n % 3 == 0)
            return false;
        for (int i = 5; i * i <= n; i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;
        return true;
    }

    static long gcd(long a, long b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

    static long fast_pow(long base, long n, long M) {
        if (n == 0)
            return 1;
        if (n == 1)
            return base % M;
        long halfn = fast_pow(base, n / 2, M);
        if (n % 2 == 0)
            return (halfn * halfn) % M;
        else
            return (((halfn * halfn) % M) * base) % M;
    }

    static long modInverse(long n, long M) {
        return fast_pow(n, M - 2, M);
    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

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

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

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