Problem: You are given an integer L. Construct a directed graph that satisfies the conditions below. The graph may contain multiple edges between the same pair of vertices. It can be proved that such a graph always exists.

* The number of vertices, N, is at most 20. The vertices are given ID numbers from 1 to N.
* The number of edges, M, is at most 60. Each edge has an integer length between 0 and 10^6 (inclusive).
* Every edge is directed from the vertex with the smaller ID to the vertex with the larger ID. That is, 1,2,...,N is one possible topological order of the vertices.
* There are exactly L different paths from Vertex 1 to Vertex N. The lengths of these paths are all different, and they are integers between 0 and L-1.



Here, the length of a path is the sum of the lengths of the edges contained in that path, and two paths are considered different when the sets of the edges contained in those paths are different.

Constraints

* 2 \leq L \leq 10^6
* L is an integer.

Input

Input is given from Standard Input in the following format:


L


Output

In the first line, print N and M, the number of the vertices and edges in your graph. In the i-th of the following M lines, print three integers u_i,v_i and w_i, representing the starting vertex, the ending vertex and the length of the i-th edge. If there are multiple solutions, any of them will be accepted.

Examples

Input

4


Output

8 10
1 2 0
2 3 0
3 4 0
1 5 0
2 6 0
3 7 0
4 8 0
5 6 1
6 7 1
7 8 1


Input

5


Output

5 7
1 2 0
2 3 1
3 4 0
4 5 0
2 4 0
1 3 3
3 5 1

Solution (in JAVA):

Solution:
import java.io.IOException;

import java.io.BufferedReader;
import java.io.InputStreamReader;

import java.io.PrintWriter;

import java.util.Arrays;
import java.util.ArrayList;

class Edge implements Comparable<Edge> {
    // 頂点番号 最小コスト
    int to;
    int cost;

    Edge(int to, int cost) {
        this.to = to;
        this.cost = cost;
    }

    // x.compereTo(y)
    // x は cost
    // y は Edge.cost
    @Override
    public int compareTo(Edge edge) {
        return this.cost - edge.cost;
    }
}

@SuppressWarnings("unchecked")
public class Main {
    static final String sp = " ";

    public static void main(String[] args) throws IOException {
        final String s;

        try (
            BufferedReader reader = new BufferedReader(
                new InputStreamReader(System.in))) {
            s = reader.readLine();
        }

        PrintWriter out = new PrintWriter(System.out);

        int L = Integer.parseInt(s);

        // log 2
        int r = (int)Math.floor(Math.log(L) / Math.log(2.0));

        int N = r + 1;

        ArrayList<Edge>[] routeList = new ArrayList[N];

        for (int i = 0; i < N; i++) routeList[i] = new ArrayList<>();

        int cnt = 0;

        // t = {0, 1, …, N - 1}
        // 頂点 t + 1 から t + 2 へ
        // コスト 2^t と 0 の辺を張る
        for (int t = 0; t < N - 1; t++) {
            // 0-indexed
            routeList[t].add(new Edge(t + 1, (int)Math.pow(2,t)));
            routeList[t].add(new Edge(t + 1, 0));
            cnt++;
            cnt++;
        }

        // t = {N - 1, N - 2, …, 1}
        // 2^r <= L - 2^(t-1) 満たすなら
        // 頂点 t から 頂点 N に対してコスト L - 2^(t-1) の辺を張る
        // L から 2^(t-1)を引く
        for (int t = N - 1; 0 < t; t--) {
            int num1 = (int)Math.pow(2, r);

            int num2 = (int)Math.pow(2, t - 1);
            if (num1 - 1 < L - num2) {
                // 0-indexed
                routeList[t - 1].add(new Edge(N - 1, L - num2));
                cnt++;
                L -= num2;
            }
        }

        // 頂点の数 辺の数
        out.println(N + sp + cnt);
        for (int i = 0; i < N; i++) {
            for (Edge e : routeList[i]) {
                // 0-indexed
                Integer from = i + 1;
                int to = e.to + 1;
                int cost = e.cost;
                out.println(from + sp + to + sp + cost);
            }
        }

        out.flush();
    }
}