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;
vector<vector<int>> res;
void f(int n) {
  if (n == 3) {
    res.push_back({1, 2, 3});
    res.push_back({1, 2, 3});
    return;
  }
  if (n == 4) {
    res.push_back({1, 2, 3});
    res.push_back({2, 3, 4});
    res.push_back({3, 4, 1});
    res.push_back({4, 1, 2});
    return;
  }
  res.push_back({n - 2, n - 1, n});
  res.push_back({n - 2, n - 1, n});
  int t = n - 3;
  while (t) {
    if (t == 3) {
      res.push_back({1, n - 1, 2, n});
      res.push_back({2, n - 1, 3, n});
      res.push_back({3, n - 1, 1, n});
      break;
    }
    res.push_back({t - 1, n - 1, t, n});
    res.push_back({t - 1, n - 1, t, n});
    t -= 2;
  }
  f(n - 2);
}
int main() {
  int n;
  scanf("%d", &n);
  f(n);
  printf("%d\n", (int)res.size());
  for (auto& v : res) {
    printf("%d", (int)v.size());
    for (int x : v) printf(" %d", x);
    puts("");
  }
  return 0;
}