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

//高速化 
struct ponjuice{ponjuice(){cin.tie(0);ios::sync_with_stdio(0);cout<<fixed<<setprecision(20);}}PonJuice;
//#define endl '\n' //インタラクティブ問題の時は消す

//型
using ll = long long;
using ld = long double;
template<class T>using vc = vector<T>; template<class T>using vvc = vc<vc<T>>; template<class T>using vvvc = vvc<vc<T>>;
using vi = vc<int>;  using vvi = vvc<int>;  using vvvi = vvvc<int>;
using vl = vc<ll>;   using vvl = vvc<ll>;   using vvvl = vvvc<ll>;
using pi = pair<int, int>;  using pl = pair<ll, ll>;
using ull = unsigned ll;
template<class T>using priq = priority_queue<T>;
template<class T>using priqg = priority_queue<T, vc<T>, greater<T>>;

// for文
#define overload4(a, b, c, d, e, ...) e
#define rep1(n)             for(ll i = 0; i < n; i++)
#define rep2(i, n)          for(ll i = 0; i < n; i++)
#define rep3(i, a, b)       for(ll i = a; i < b; i++)
#define rep4(i, a, b, step) for(ll i = a; i < b; i+= step)
#define rep(...) overload4(__VA_ARGS__, rep4, rep3, rep2, rep1)(__VA_ARGS__)
#define per1(n)             for(ll i = n-1; i >= 0; i--)
#define per2(i, n)          for(ll i = n-1; i >= 0; i--)
#define per3(i, a, b)       for(ll i = b-1; i >= a; i--)
#define per4(i, a, b, step) for(ll i = b-1; i >= a; i-= step)
#define per(...) overload4(__VA_ARGS__, per4, per3, per2, per1)(__VA_ARGS__)

//関数
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define si(x) (ll)(x).size()
template<class S, class T>inline bool chmax(S& a, T b){return a < b && ( a = b , true);}
template<class S, class T>inline bool chmin(S& a, T b){return a > b && ( a = b , true);}

inline void yes(){cout << "Yes\n";}
inline void no(){cout << "No\n";}
inline void yesno(bool y = true){if(y)yes();else no();}

//定数
constexpr ll mod = 998244353;
constexpr ll minf=-(1<<29);
constexpr ll inf=(1<<29);
constexpr ll MINF=-(1LL<<60);
constexpr ll INF=(1LL<<60);
const int dx[4] ={-1, 0, 1, 0};
const int dy[4] ={ 0, 1, 0,-1};
const int dx8[8] ={-1,-1,-1, 0, 1, 1, 1, 0};
const int dy8[8] ={-1, 0, 1, 1, 1, 0,-1,-1};

void solve();
int main() {
	int t = 1;
    // cin >>t;
    while(t--){
        solve();
    }
}

/*
せっかくなので、PPC は全てコメント書きます


構築いいね👍

どうせ1とか4とかがすべてになる

1 - 2 を1
1 - i (i>=3) を 1,3 でもつ
と全ての頂点対について1か4でいけるね

全ての辺の重みは相異なる 1 以上 200 以下の整数である
をみていなかった... バカ


ほなむずいか

1 3 5 7 9
の累積和がi^2になることを使いたくて、頑張るしかないね

(i) - (i+1) に i*2-1 をつけて
(2) - i に i*2-2 をつけてあげると良さそう
*/

void solve(){
    ll n;
    cin >> n;

    cout << n * 2 - 3 << endl;
    rep(i,0,n-1) {
        cout << i+1 << " " << i+2 << " " << i*2+1 << endl;
    }
    rep(i,2,n) {
        cout << 2 << " " << i+1 << " " << i*2 << endl;
    }
    rep(i,0,n){
        rep(j,i+1,n) {
            cerr << "(" << i+1 << " " << j+1 <<") ";
            if(i == 0) {
                cout << j << " ";
                rep(k,0,j) {
                    cout << k+1 << " ";
                }cout << endl;
                continue;
            }
            if(i == 1) {
                cout << j-1 << " ";
                cout << n << " ";
                rep(k,0,j-2) {
                    cout << k+3 << " ";
                }cout << endl;
                continue;
            }
            // 2まで戻ってから一手で行く
            cout << j << " ";
            per(k,2,i+1) {
                cout << k << " ";
            }
            cout << n + i-1 << " ";
            rep(k,i+1,j) {
                cout << k+1 << " ";
            }
            cout << endl;
        }
    }
}