Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

4

Sample Output 1

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

In the graph represented by the sample output, there are four paths from Vertex 1 to N=8:

1 → 2 → 3 → 4 → 8 with length 0

1 → 2 → 3 → 7 → 8 with length 1

1 → 2 → 6 → 7 → 8 with length 2

1 → 5 → 6 → 7 → 8 with length 3

There are other possible solutions.

Sample Input 2

5

Sample Output 2

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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;

// Optimization

//#pragma GCC optimize("O3")
#define endl '\n'

// Shortcut

//#define int long long
#define eb emplace_back
#define pb push_back
#define pob pop_back
#define mp make_pair
#define upb upper_bound
#define lwb lower_bound
#define fi first
#define se second
#define For(i, l, r) for (int i = l; i < r; i++)
#define ForE(i, l, r) for (int i = l; i <= r; i++)
#define Ford(i, r, l) for (int i = r; i > l; i--)
#define FordE(i, r, l) for (int i = r; i >= l; i--)
#define Fora(i, a) for (auto i : a)

// I/O & Debug

#define PrintV(a) Fora(iiii, a) cout << iiii << ' '; cout << endl;
#define PrintVl(a) Fora(iiii, a) cout << iiii << endl;
#define PrintA(a, l, r) for (int iiii = l; iiii <= r; iiii++) cout << a[iiii] << ' '; cout << endl;
#define PrintAl(a, l, r) for (int iiii = l; iiii <= r; iiii++) cout << a[iiii] << endl;
#define Ptest(x) return cout << x, 0;
#define gl(s) getline(cin, s);
#define setpre(x) fixed << setprecision(x)
/*
#define debug(args...){ string _sDEB = #args; replace(_sDEB.begin(), _sDEB.end(), ',', ' '); stringstream _ssDEB(_sDEB); istream_iterator<string> _itDEB(_ssDEB); DEB(_itDEB, args); }
void DEB(istream_iterator<string> it) {}
template<typename T, typename... Args>
void DEB(istream_iterator<string> it, T a, Args... args){
    cout << *it << " = " << a << endl;
    DEB(++it, args...);
}
*/

// Functions

//#define isvowel(a) (a == 'a' || a == 'e' || a == 'i' || a == 'o' || a == 'u')
#define bend(a) a.begin(), a.end()
#define rbend(a) a.rbegin(), a.rend()
#define mset(a) memset(a, 0, sizeof(a))
#define mset1(a) memset(a, 1, sizeof(a))
#define msetn1(a) memset(a, -1, sizeof(a))
#define msetinf(a) memset(a, 0x3f, sizeof(a))
#define gcd __gcd
#define __builtin_popcount __builtin_popcountll
//mt19937 rando(chrono::steady_clock::now().time_since_epoch().count());

// Data Structure

#define pque priority_queue
#define mts multiset
#define y0 _y0_
#define y1 _y1_
#define div divi
typedef long long ll;
typedef long double ld;
typedef vector <int> vi;
typedef vector <ll> vll;
typedef vector <ld> vld;
typedef vector <string> vs;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef vector <vi > vvi;
typedef vector <vll > vvll;
typedef vector <pii > vpii;
typedef vector <pll > vpll;

const int N = 20 + 5, mod = 1e9 + 7, mod1 = 998244353, mod2 = 1e9 + 9, inf = 1e9 + 7;
const ll infll = 1e18 + 7;

int l;
vpii adj[N];
vector <pair <pii, int>> edge;

signed main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    cin >> l;
    int l2 = log2(l), l3 = l;
    ForE(i, 2, l2 + 1){
    	adj[i - 1].pb({i, 0});
    	adj[i - 1].pb({i, (1 << (i - 2))});
	}
	For(i, 0, l2){
		if (l & (1 << i)){
			adj[i + 1].pb({l2 + 1, l3 - (1 << i)});
			l3 -= (1 << i);
		}
	}
	ForE(i, 1, l2 + 1){
		Fora(v, adj[i]){
			edge.pb({{i, v.fi}, v.se});
		}
	}
	cout << l2 + 1 << ' ' << edge.size() << endl;
	Fora(v, edge){
		cout << v.fi.fi << ' ' << v.fi.se << ' ' << v.se << endl;
	}
}

/*
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*/