1. 先从结论说起，首先一条线的斜率 (y - 0) / (x - 0) = y / x，同样斜率的线只会有一个点，其它的点都会被这个点给挡住
2. 这个点的y 和 x 肯定是互质的 gcd(x, y) = 1
3. 欧拉函数搞一搞，有3个特殊点，其他都 res + E[2 : N] * 2

Java 代码

// Don't place your source in a package


import javax.swing.*;
import java.lang.reflect.Array;
import java.text.DecimalFormat;
import java.util.*;
import java.lang.*;
import java.io.*;
import java.math.*;
import java.util.stream.Stream;



// Please name your class Main
public class Main {
    static FastScanner fs=new FastScanner();
    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        public String next() {
            while (!st.hasMoreElements())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }
        int Int() {
            return Integer.parseInt(next());
        }

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

        String Str(){
            return next();
        }
    }


    public static void main (String[] args) throws java.lang.Exception {
        PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        //reading /writing file
        //Scanner sc=new Scanner(new File("input.txt"));
        //PrintWriter pr=new PrintWriter("output.txt");



        int T=Int();
        for(int t=0;t<T;t++){
            int n=Int();

            Solution sol=new Solution(out);
            sol.solution(n,t);
        }
        out.close();

    }


    public static int[] Arr(int n){
        int A[]=new int[n];
        for(int i=0;i<n;i++)A[i]=Int();
        return A;
    }
    public static int Int(){
        return fs.Int();
    }
    public static long Long(){
        return fs.Long();
    }
    public static String Str(){
        return fs.Str();
    }

}



class Solution {
    PrintWriter out;
    int INF = Integer.MAX_VALUE;
    int MOD = 1000000009;
    int mod = MOD;
    public Solution(PrintWriter out) {
        this.out = out;
    }




    public void solution(int N,int t) {
        int res=3;
        int E[]=euler(N+10);
        for(int i=2;i<=N;i++)res+=(E[i]*2);

        out.println((t+1)+" "+N+" "+res);
    }

    public int[] euler(int n) {
        int E[]=new int[n+1];
        E[1]=1;
        for(int i=2;i<E.length;i++)
            E[i]=i;
        for(int i=2;i<E.length;i++){
            if(E[i]==i)
            {
                for(int j=i;j<E.length;j+=i){
                    E[j]=E[j]/i*(i-1);
                }
            }
        }
        return E;
    }



}