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;
const int mod = 1e9 + 7;
vector<tuple<int, int, int>> jaji;
vector<tuple<int, int, int, int>> boji;
int piv;
void add3cyc(int x, int y, int z) { jaji.emplace_back(x, y, z); }
void add4cyc(int x, int y, int z, int w) { boji.emplace_back(x, y, z, w); }
vector<int> solve(int x) {
  if (x == 2) {
    vector<int> v;
    for (int i = 0; i < 2; i++) v.push_back(++piv);
    return v;
  }
  if (x == 3) {
    vector<int> v;
    for (int i = 0; i < 3; i++) {
      v.push_back(++piv);
    }
    for (int i = 0; i < 2; i++) add3cyc(v[0], v[1], v[2]);
    return v;
  }
  if (x == 4) {
    vector<int> v;
    for (int i = 0; i < 4; i++) {
      v.push_back(++piv);
    }
    add3cyc(v[0], v[1], v[2]);
    add3cyc(v[0], v[1], v[3]);
    add3cyc(v[0], v[2], v[3]);
    add3cyc(v[1], v[2], v[3]);
    return v;
  }
  if (x == 5) {
    vector<int> v;
    for (int i = 0; i < 5; i++) v.push_back(++piv);
    add3cyc(v[4], v[3], v[1]);
    add3cyc(v[2], v[0], v[4]);
    add4cyc(v[3], v[4], v[1], v[2]);
    add4cyc(v[3], v[2], v[1], v[0]);
    add3cyc(v[3], v[1], v[0]);
    add3cyc(v[2], v[0], v[4]);
    return v;
  }
  if (x == 6) {
    vector<int> a = solve(2);
    vector<int> b = solve(2);
    vector<int> c = solve(2);
    add3cyc(a[0], a[1], b[0]);
    add3cyc(a[0], a[1], b[1]);
    add3cyc(b[0], b[1], c[0]);
    add3cyc(b[0], b[1], c[1]);
    add3cyc(c[0], c[1], a[0]);
    add3cyc(c[0], c[1], a[1]);
    add4cyc(a[0], b[0], a[1], b[1]);
    add4cyc(b[0], c[0], b[1], c[1]);
    add4cyc(c[0], a[0], c[1], a[1]);
    vector<int> r;
    for (auto &v : {a, b, c}) {
      for (auto &w : v) r.push_back(w);
    }
    return r;
  }
  vector<int> l, r;
  if (x % 2 == 1) {
    if (x % 4 == 3) {
      l = solve(x / 2 + 1);
      r = solve(x / 2);
    } else {
      l = solve(x / 2);
      r = solve(x / 2 + 1);
    }
    for (int i = 0; i < l.size(); i += 2) {
      for (int j = 0; j < 2 * r.size(); j += 2) {
        add4cyc(l[i], r[j % r.size()], l[i + 1], r[(j + 1) % r.size()]);
      }
    }
  } else {
    if (x % 4 == 2) {
      l = solve(x / 2 - 1);
      r = solve(x / 2 + 1);
    } else {
      l = solve(x / 2);
      r = solve(x / 2);
    }
    for (int i = 0; i < l.size(); i += 2) {
      for (int j = 0; j < r.size(); j += 2) {
        for (int k = 0; k < 2; k++) add4cyc(l[i], r[j], l[i + 1], r[j + 1]);
      }
    }
  }
  vector<int> v;
  for (auto &i : l) v.push_back(i);
  for (auto &i : r) v.push_back(i);
  return v;
}
int main() {
  int n;
  cin >> n;
  solve(n);
  cout << jaji.size() + boji.size() << endl;
  for (auto &i : jaji) printf("3 %d %d %d\n", get<0>(i), get<1>(i), get<2>(i));
  for (auto &i : boji)
    printf("4 %d %d %d %d\n", get<0>(i), get<1>(i), get<2>(i), get<3>(i));
}