#include <bits/stdc++.h>
const int maxn = 300, maxm = maxn * maxn;
int n, m, out[maxm | 1][5];
void addEdge(int a, int b, int c) {
  out[m][0] = 3;
  out[m][1] = a;
  out[m][2] = b;
  out[m++][3] = c;
}
void addEdge(int a, int b, int c, int d) {
  out[m][0] = 4;
  out[m][1] = a;
  out[m][2] = b;
  out[m][3] = c;
  out[m++][4] = d;
}
void showEdge() {
  printf("%d\n", m);
  for (int i = 0; i < m; ++i)
    for (int j = 0; j <= out[i][0]; ++j)
      printf("%d%c", out[i][j], " \n"[j == out[i][0]]);
}
int main() {
  scanf("%d", &n);
  if (n & 1) {
    for (int i = 3; i <= n; i += 2) {
      addEdge(1, i - 1, i);
      addEdge(1, i - 1, i);
      for (int j = 3; j < i; j += 2) {
        addEdge(j - 1, i, j, i - 1);
        addEdge(j - 1, i, j, i - 1);
      }
    }
  } else {
    addEdge(1, 2, 3, 4);
    addEdge(1, 2, 4, 3);
    addEdge(1, 3, 2, 4);
    for (int i = 6; i <= n; i += 2) {
      addEdge(1, i - 1, i);
      addEdge(2, i - 1, i);
      addEdge(1, i, 2, i - 1);
      for (int j = 4; j < i; j += 2) {
        addEdge(j - 1, i, j, i - 1);
        addEdge(j - 1, i, j, i - 1);
      }
    }
  }
  showEdge();
  return 0;
}

### Prompt

Create a solution in Cpp for the following 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

### Response

```cpp
#include <bits/stdc++.h>
const int maxn = 300, maxm = maxn * maxn;
int n, m, out[maxm | 1][5];
void addEdge(int a, int b, int c) {
  out[m][0] = 3;
  out[m][1] = a;
  out[m][2] = b;
  out[m++][3] = c;
}
void addEdge(int a, int b, int c, int d) {
  out[m][0] = 4;
  out[m][1] = a;
  out[m][2] = b;
  out[m][3] = c;
  out[m++][4] = d;
}
void showEdge() {
  printf("%d\n", m);
  for (int i = 0; i < m; ++i)
    for (int j = 0; j <= out[i][0]; ++j)
      printf("%d%c", out[i][j], " \n"[j == out[i][0]]);
}
int main() {
  scanf("%d", &n);
  if (n & 1) {
    for (int i = 3; i <= n; i += 2) {
      addEdge(1, i - 1, i);
      addEdge(1, i - 1, i);
      for (int j = 3; j < i; j += 2) {
        addEdge(j - 1, i, j, i - 1);
        addEdge(j - 1, i, j, i - 1);
      }
    }
  } else {
    addEdge(1, 2, 3, 4);
    addEdge(1, 2, 4, 3);
    addEdge(1, 3, 2, 4);
    for (int i = 6; i <= n; i += 2) {
      addEdge(1, i - 1, i);
      addEdge(2, i - 1, i);
      addEdge(1, i, 2, i - 1);
      for (int j = 4; j < i; j += 2) {
        addEdge(j - 1, i, j, i - 1);
        addEdge(j - 1, i, j, i - 1);
      }
    }
  }
  showEdge();
  return 0;
}

```