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.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Map;

/**
 * Copyright © 2018 Chris. All rights reserved.
 *
 * @author Chris
 * 2018/7/9 15:33
 * @see format
 */
public class C {

    private static BufferedReader br;
    private static StreamTokenizer st;
    private static PrintWriter pw;

    static final int INF = 1000000007;
    static final int MOD = 1000000007;

    static int n, m, T;
    static long num;
    static int[] a, b, c;
    static long[] al, bl, cl;
    static String s, t;
    static long C[];

    private static void solve() throws IOException {
        int n = nextInt();
        long a[] = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = nextInt();
        }
        pw.println(n + 1);
        long add = 0;
        for (int i = n - 1; i >= 0; i--) {
            long cur = add + a[i];
            long mod = cur % n;
            add += (i - mod);
            pw.format("1 %d %d\n", i + 1, (i - mod) + n);
        }
        pw.format("2 %d %d", n, n);
    }

    static void swap(long a[], int i, int j) {
        a[i] ^= a[j];
        a[j] ^= a[i];
        a[i] ^= a[j];
    }

    static void swap(int a[], int i, int j) {
        a[i] ^= a[j];
        a[j] ^= a[i];
        a[i] ^= a[j];
    }

    static void getDiv(Map<Integer, Integer> map, int n) {
        int sqrt = (int) Math.sqrt(n);
        for (int i = sqrt; i >= 2; i--) {
            if (n % i == 0) {
                getDiv(map, i);
                getDiv(map, n / i);
                return;
            }
        }
        map.put(n, map.getOrDefault(n, 0) + 1);
    }

    public static boolean[] generatePrime(int n) {
        boolean p[] = new boolean[n + 1];
        p[2] = true;

        for (int i = 3; i <= n; i += 2) {
            p[i] = true;
        }

        for (int i = 3; i <= Math.sqrt(n); i += 2) {
            if (!p[i]) {
                continue;
            }
            for (int j = i * i; j <= n; j += i << 1) {
                p[j] = false;
            }
        }
        return p;
    }

    // 18位素数：154590409516822759
    // 19位素数：2305843009213693951 (梅森素数)
    // 19位素数：4384957924686954497
    static boolean isPrime(long n) { //determines if n is a prime number
        int p[] = {2, 3, 5, 233, 331};
        int pn = p.length;
        long s = 0, t = n - 1;//n - 1 = 2^s * t
        while ((t & 1) == 0) {
            t >>= 1;
            ++s;
        }
        for (int i = 0; i < pn; ++i) {
            if (n == p[i]) {
                return true;
            }
            long pt = pow(p[i], t, n);
            for (int j = 0; j < s; ++j) {
                long cur = llMod(pt, pt, n);
                if (cur == 1 && pt != 1 && pt != n - 1) {
                    return false;
                }
                pt = cur;
            }
            if (pt != 1) {
                return false;
            }
        }
        return true;
    }

    static long llMod(long a, long b, long mod) {
        return (a * b - (long) ((double) a / mod * b + 0.5) * mod + mod) % mod;
//        long r = 0;
//        a %= mod;
//        b %= mod;
//        while (b > 0) {
//            if ((b & 1) == 1) {
//                r = (r + a) % mod;
//            }
//            b >>= 1;
//            a = (a << 1) % mod;
//        }
//        return r;
    }

    static int pow(long a, long n) {
        long ans = 1;
        while (n > 0) {
            if ((n & 1) == 1) {
                ans = (ans * a) % MOD;
            }
            a = (a * a) % MOD;
            n >>= 1;
        }
        return (int) ans;
    }

    static int pow(long a, long n, long mod) {
        long ans = 1;
        while (n > 0) {
            if ((n & 1) == 1) {
                ans = llMod(ans, a, mod);
            }
            a = llMod(a, a, mod);
            n >>= 1;
        }
        return (int) ans;
    }

    private static long[][] initC(int n) {
        long c[][] = new long[n][n];

        for (int i = 0; i < n; i++) {
            c[i][0] = 1;
        }

        for (int i = 1; i < n; i++) {
            for (int j = 1; j <= i; j++) {
                c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
            }
        }
        return c;
    }

    /**
     * ps: n >= m, choose m from n;
     */
    private static int c(long n, long m) {
        if (m > n) {
            n ^= m;
            m ^= n;
            n ^= m;
        }
        m = Math.min(m, n - m);

        long top = 1;
        long bot = 1;
        for (long i = n - m + 1; i <= n; i++) {
            top = (top * i) % MOD;
        }
        for (int i = 1; i <= m; i++) {
            bot = (bot * i) % MOD;
        }

        return (int) ((top * pow(bot, MOD - 2)) % MOD);
    }

    static int gcd(int a, int b) {
        if (a < b) {
            a ^= b;
            b ^= a;
            a ^= b;
        }
        while (b != 0) {
            int tmp = a % b;
            a = b;
            b = tmp;
        }
        return a;
    }

    static boolean even(long n) {
        return (n & 1) == 0;
    }

    public static void main(String args[]) throws IOException {
        boolean oj = System.getProperty("ONLINE_JUDGE") != null;
        if (!oj) {
            System.setIn(new FileInputStream("in.txt"));
//            System.setOut(new PrintStream("out.txt"));
        }
        br = new BufferedReader(new InputStreamReader(System.in));
        st = new StreamTokenizer(br);
        pw = new PrintWriter(new OutputStreamWriter(System.out));
        st.ordinaryChar('\''); //指定单引号、双引号和注释符号是普通字符
        st.ordinaryChar('\"');
        st.ordinaryChar('/');

        long t = System.currentTimeMillis();
        solve();
        if (!oj) {
            pw.println("[" + (System.currentTimeMillis() - t) + "ms]");
        }
        pw.flush();
    }

    private static long[] anLong(int n) throws IOException {
        long a[] = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = nextInt();
        }
        return a;
    }

    private static String next(int len) throws IOException {
        st.nextToken();
        return st.sval;
    }

    private static int nextInt() throws IOException {
        st.nextToken();
        return (int) st.nval;
    }

    private static long nextLong() throws IOException {
        return Long.parseLong(nextLine());
    }

    private static double nextDouble() throws IOException {
        st.nextToken();
        return st.nval;
    }

    private static String[] nextSS(String reg) throws IOException {
        return br.readLine().split(reg);
    }

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