#include <bits/stdc++.h>
using namespace std;
const long long int N = 5e6 + 10, N2 = 2e5, delta = 10501, mod = 998244353,
                    oo = 1e18, LOG = 20, SQ = 300;
const long double PI = 3.141592653589793;
long long int sz;
vector<long long int> v[N];
void dor3(long long int a, long long int b, long long int c) {
  v[sz] = {a, b, c};
}
void dor4(long long int a, long long int b, long long int c, long long int d) {
  v[sz] = {a, b, c, d};
}
void solve(long long int n) {
  if (n == 3) {
    dor3(1, 2, 3);
    sz++;
    dor3(1, 2, 3);
    sz++;
    return;
  }
  if (n == 4) {
    dor4(1, 2, 3, 4);
    sz++;
    dor4(1, 3, 4, 2);
    sz++;
    dor4(1, 4, 2, 3);
    sz++;
    return;
  }
  solve(n - 2);
  if (n & 1) {
    dor3(1, n - 1, n);
    sz++;
    dor3(1, n - 1, n);
    sz++;
    for (long long int i = 2; i < n - 2; i += 2) {
      dor4(i, n - 1, i + 1, n);
      sz++;
      dor4(i, n - 1, i + 1, n);
      sz++;
    }
    return;
  }
  dor4(1, n - 1, 2, n);
  sz++;
  dor3(1, n - 1, n);
  sz++;
  dor3(2, n - 1, n);
  sz++;
  for (long long int i = 3; i < n - 2; i += 2) {
    dor4(i, n - 1, i + 1, n);
    sz++;
    dor4(i, n - 1, i + 1, n);
    sz++;
  }
}
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long int n;
  cin >> n;
  solve(n);
  cout << sz << '\n';
  for (long long int i = 0; i < sz; i++) {
    cout << v[i].size() << ' ';
    for (long long int u : v[i]) cout << u << ' ';
    cout << '\n';
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long int N = 5e6 + 10, N2 = 2e5, delta = 10501, mod = 998244353,
                    oo = 1e18, LOG = 20, SQ = 300;
const long double PI = 3.141592653589793;
long long int sz;
vector<long long int> v[N];
void dor3(long long int a, long long int b, long long int c) {
  v[sz] = {a, b, c};
}
void dor4(long long int a, long long int b, long long int c, long long int d) {
  v[sz] = {a, b, c, d};
}
void solve(long long int n) {
  if (n == 3) {
    dor3(1, 2, 3);
    sz++;
    dor3(1, 2, 3);
    sz++;
    return;
  }
  if (n == 4) {
    dor4(1, 2, 3, 4);
    sz++;
    dor4(1, 3, 4, 2);
    sz++;
    dor4(1, 4, 2, 3);
    sz++;
    return;
  }
  solve(n - 2);
  if (n & 1) {
    dor3(1, n - 1, n);
    sz++;
    dor3(1, n - 1, n);
    sz++;
    for (long long int i = 2; i < n - 2; i += 2) {
      dor4(i, n - 1, i + 1, n);
      sz++;
      dor4(i, n - 1, i + 1, n);
      sz++;
    }
    return;
  }
  dor4(1, n - 1, 2, n);
  sz++;
  dor3(1, n - 1, n);
  sz++;
  dor3(2, n - 1, n);
  sz++;
  for (long long int i = 3; i < n - 2; i += 2) {
    dor4(i, n - 1, i + 1, n);
    sz++;
    dor4(i, n - 1, i + 1, n);
    sz++;
  }
}
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long int n;
  cin >> n;
  solve(n);
  cout << sz << '\n';
  for (long long int i = 0; i < sz; i++) {
    cout << v[i].size() << ' ';
    for (long long int u : v[i]) cout << u << ' ';
    cout << '\n';
  }
  return 0;
}

```