Problem: 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].

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;

public class Today {
	public static void main(String[] args) throws IOException, InterruptedException {
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		StringBuilder sb = new StringBuilder();
		int n = sc.nextInt();
		long sum = 0;
		if (n == 1) {
			System.out.println(0);
			return;
		}
		int a[] = new int[n];
		int c = 0;
		for (int i = 0; i < n; i++)
			a[i] = sc.nextInt();
		if (a[n - 1] < n) {
			sb.append(1 + " " + n + " " + (n - a[n - 1]) + "\n");
			sum += (n - a[n - 1]);
			c++;
		}
		for (int i = n - 2; i >= 0; i--) {
			long remainder = (a[i] + sum) % n;
			remainder = n - remainder;
			remainder += (i + 1);
			sb.append(1 + " " + (i + 1) + " " + remainder + "\n");
			sum += remainder;
			c++;
		}
		sb.append(2 + " " + (n - 1) + " " + n);
		pw.println(c + 1);
		pw.println(sb.toString());
		pw.flush();
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

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

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

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

		public boolean ready() throws IOException {
			return br.ready();
		}

		public void waitForInput() throws InterruptedException {
			Thread.sleep(4000);
		}
	}
}