Problem: In order to fly to the Moon Mister B just needs to solve the following problem.

There is a complete indirected graph with n vertices. You need to cover it with several simple cycles of length 3 and 4 so that each edge is in exactly 2 cycles.

We are sure that Mister B will solve the problem soon and will fly to the Moon. Will you?

Input

The only line contains single integer n (3 ≤ n ≤ 300).

Output

If there is no answer, print -1.

Otherwise, in the first line print k (1 ≤ k ≤ n2) — the number of cycles in your solution.

In each of the next k lines print description of one cycle in the following format: first print integer m (3 ≤ m ≤ 4) — the length of the cycle, then print m integers v1, v2, ..., vm (1 ≤ vi ≤ n) — the vertices in the cycle in the traverse order. Each edge should be in exactly two cycles.

Examples

Input

3


Output

2
3 1 2 3
3 1 2 3


Input

5


Output

6
3 5 4 2
3 3 1 5
4 4 5 2 3
4 4 3 2 1
3 4 2 1
3 3 1 5

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
constexpr int MOD = 1000000007;
template <typename T, typename U>
std::istream& operator>>(std::istream& i, pair<T, U>& p) {
  i >> p.first >> p.second;
  return i;
}
template <typename T>
std::istream& operator>>(std::istream& i, vector<T>& t) {
  for (auto& v : t) {
    i >> v;
  }
  return i;
}
template <typename T>
std::ostream& operator<<(std::ostream& o, const vector<T>& t) {
  for (size_t i = 0; i < t.size(); ++i) {
    o << t[i] << " \n"[i == t.size() - 1];
  }
  return o;
}
template <typename T>
using minheap = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
using maxheap = priority_queue<T, vector<T>, less<T>>;
auto fraclt = [](const std::pair<int, int>& a, const std::pair<int, int>& b) {
  return (long long)a.first * b.second < (long long)b.first * a.second;
};
struct cmpfrac {
  bool operator()(const std::pair<int, int>& a,
                  const std::pair<int, int>& b) const {
    return (long long)a.first * b.second < (long long)b.first * a.second;
  }
};
template <typename T>
bool in(T a, T b, T c) {
  return a <= b && b < c;
}
int logceil(long long first) {
  int b = 0;
  while (first) {
    first >>= 1;
    ++b;
  }
  return b;
}
namespace std {
template <typename T, typename U>
struct hash<pair<T, U>> {
  hash<T> t;
  hash<U> u;
  size_t operator()(const pair<T, U>& p) const {
    return t(p.first) ^ (u(p.second) << 7);
  }
};
}  // namespace std
template <typename T, typename F>
T bsh(T l, T h, const F& f) {
  T r = -1, m;
  while (l <= h) {
    m = (l + h) / 2;
    if (f(m)) {
      l = m + 1;
      r = m;
    } else {
      h = m - 1;
    }
  }
  return r;
}
template <typename T, typename F>
T bsl(T l, T h, const F& f) {
  T r = -1, m;
  while (l <= h) {
    m = (l + h) / 2;
    if (f(m)) {
      h = m - 1;
      r = m;
    } else {
      l = m + 1;
    }
  }
  return r;
}
template <typename T>
T gcd(T a, T b) {
  if (a < b) swap(a, b);
  return b ? gcd(b, a % b) : a;
}
template <typename T>
class vector2 : public vector<vector<T>> {
 public:
  vector2() {}
  vector2(size_t a, size_t b, T t = T())
      : vector<vector<T>>(a, vector<T>(b, t)) {}
};
template <typename T>
class vector3 : public vector<vector<vector<T>>> {
 public:
  vector3() {}
  vector3(size_t a, size_t b, size_t c, T t = T())
      : vector<vector<vector<T>>>(a, vector<vector<T>>(b, vector<T>(c, t))) {}
};
template <typename T>
struct bounded_priority_queue {
  inline bounded_priority_queue(unsigned int X) : A(X), B(0) {}
  inline void push(unsigned int L, T V) {
    B = max(B, L);
    A[L].push(V);
  }
  inline const T& top() const { return A[B].front(); }
  inline void pop() {
    A[B].pop();
    while (B > 0 && A[B].empty()) --B;
  }
  inline bool empty() const { return A[B].empty(); }
  inline void clear() {
    B = 0;
    for (auto& a : A) a = queue<T>();
  }

 private:
  vector<queue<T>> A;
  unsigned int B;
};
class TaskE {
 public:
  void solve(istream& cin, ostream& cout) {
    int N;
    cin >> N;
    vector<vector<int>> A;
    if (N == 4) {
      A.push_back({1, 2, 3});
      A.push_back({1, 2, 4});
      A.push_back({1, 4, 3});
      A.push_back({4, 2, 3});
    } else {
      int k = N / 2;
      for (int i = 0; i < k; ++i) {
        for (int j = i + 1; j < k; ++j) {
          A.push_back({2 * i + 1, 2 * j + 1, 2 * i + 2, 2 * j + 2});
          if (N % 2 == 0 && (j == (i + 1) || (i == 0 && j == k - 1))) {
            if (j == i + 1) {
              A.push_back({2 * i + 1, 2 * j + 2, 2 * i + 2});
              A.push_back({2 * i + 1, 2 * j + 1, 2 * i + 2});
            } else {
              A.push_back({2 * j + 1, 2 * i + 2, 2 * j + 2});
              A.push_back({2 * j + 1, 2 * i + 1, 2 * j + 2});
            }
          } else {
            A.push_back({2 * i + 1, 2 * j + 1, 2 * i + 2, 2 * j + 2});
          }
        }
        if (N % 2) {
          A.push_back({2 * i + 1, N, 2 * i + 2});
          A.push_back({2 * i + 1, N, 2 * i + 2});
        }
      }
    }
    cout << A.size() << endl;
    for (auto& a : A) cout << (int)a.size() << ' ' << a;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  TaskE solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}