You're given an array a of length n. You can perform the following operations on it:

  * choose an index i (1 ≤ i ≤ n), an integer x (0 ≤ x ≤ 10^6), and replace a_j with a_j+x for all (1 ≤ j ≤ i), which means add x to all the elements in the prefix ending at i. 
  * choose an index i (1 ≤ i ≤ n), an integer x (1 ≤ x ≤ 10^6), and replace a_j with a_j \% x for all (1 ≤ j ≤ i), which means replace every element in the prefix ending at i with the remainder after dividing it by x. 



Can you make the array strictly increasing in no more than n+1 operations?

Input

The first line contains an integer n (1 ≤ n ≤ 2000), the number of elements in the array a.

The second line contains n space-separated integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^5), the elements of the array a.

Output

On the first line, print the number of operations you wish to perform. On the next lines, you should print the operations.

To print an adding operation, use the format "1 i x"; to print a modding operation, use the format "2 i x". If i or x don't satisfy the limitations above, or you use more than n+1 operations, you'll get wrong answer verdict.

Examples

Input

3
1 2 3


Output

0

Input

3
7 6 3


Output

2
1 1 1
2 2 4

Note

In the first sample, the array is already increasing so we don't need any operations.

In the second sample:

In the first step: the array becomes [8,6,3].

In the second step: the array becomes [0,2,3].
import java.util.*;
import java.io.*; 
import java.text.*;
//Solution Credits: Taranpreet Singh
public class Main{
    //SOLUTION BEGIN
    void solve(int TC) throws Exception{
        int n = ni();
        int[] a = new int[n];
        for(int i = 0; i< n; i++)a[i] = ni();
        int sum = 0;
        pn(n+1);
        for(int i = n-1; i>= 0; i--){
            int x = (i+n-(sum+a[i])%n)%n;
            pn("1 "+(i+1)+" " +x);
            sum+=x;
        }
        pn("2 "+n+" " +n);
    }
    void shuffle(long[] a){
        Random r = new Random();
        for(int i = 0; i< a.length; i++){
            int j = Math.abs(r.nextInt())%a.length;
            long tmp = a[i];
            a[i] = a[j];
            a[j] = tmp;
        }
    }
    //SOLUTION END
    long mod = (long)1e9+7, IINF = (long)5e18;
    final int INF = (int)2e9;
    DecimalFormat df = new DecimalFormat("0.000000000");
    double PI = 3.1415926535897932384626433832792884197169399375105820974944, eps = 1e-8;
    static boolean multipleTC = false, memory = false;
    FastReader in;PrintWriter out;
    void run() throws Exception{
        in = new FastReader();
        out = new PrintWriter(System.out);
        int T = (multipleTC)?ni():1;
        //Solution Credits: Taranpreet Singh
        for(int i = 1; i<= T; i++)solve(i);
        out.flush();
        out.close();
    }
    public static void main(String[] args) throws Exception{
        if(memory)new Thread(null, new Runnable() {public void run(){try{new Main().run();}catch(Exception e){e.printStackTrace();}}}, "1", 1 << 28).start();
        else new Main().run();
    }
    long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
    int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
    int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
    void p(Object o){out.print(o);}
    void pn(Object o){out.println(o);}
    void pni(Object o){out.println(o);out.flush();}
    String n(){return in.next();}
    String nln(){return in.nextLine();}
    int ni(){return Integer.parseInt(in.next());}
    long nl(){return Long.parseLong(in.next());}
    double nd(){return Double.parseDouble(in.next());}

    class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public FastReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }

        String next(){
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        String nextLine(){
            String str = "";
            try{    
                str = br.readLine();
            }catch (IOException e){
                e.printStackTrace();
            }   
            return str;
        }
    }
}  