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>
clock_t t = clock();
namespace my_std {
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
inline T rnd(T l, T r) {
  return uniform_int_distribution<T>(l, r)(rng);
}
template <typename T>
inline bool chkmax(T& x, T y) {
  return x < y ? x = y, 1 : 0;
}
template <typename T>
inline bool chkmin(T& x, T y) {
  return x > y ? x = y, 1 : 0;
}
template <typename T>
inline void read(T& t) {
  t = 0;
  char f = 0, ch = getchar();
  double d = 0.1;
  while (ch > '9' || ch < '0') f |= (ch == '-'), ch = getchar();
  while (ch <= '9' && ch >= '0') t = t * 10 + ch - 48, ch = getchar();
  if (ch == '.') {
    ch = getchar();
    while (ch <= '9' && ch >= '0') t += d * (ch ^ 48), d *= 0.1, ch = getchar();
  }
  t = (f ? -t : t);
}
template <typename T, typename... Args>
inline void read(T& t, Args&... args) {
  read(t);
  read(args...);
}
char __sr[1 << 21], __z[20];
int __C = -1, __zz = 0;
inline void Ot() { fwrite(__sr, 1, __C + 1, stdout), __C = -1; }
inline void print(register int x) {
  if (__C > 1 << 20) Ot();
  if (x < 0) __sr[++__C] = '-', x = -x;
  while (__z[++__zz] = x % 10 + 48, x /= 10)
    ;
  while (__sr[++__C] = __z[__zz], --__zz)
    ;
  __sr[++__C] = '\n';
}
void file() {}
inline void chktime() {}
long long ksm(long long x, int y) {
  long long ret = 1;
  for (; y; y >>= 1, x = x * x)
    if (y & 1) ret = ret * x;
  return ret;
}
}  // namespace my_std
using namespace my_std;
int n;
int ans[10][333][10] = {
    {{2}, {3, 1, 2, 3}, {3, 1, 2, 3}},
    {{4}, {3, 1, 2, 3}, {3, 1, 2, 4}, {3, 1, 3, 4}, {3, 2, 3, 4}},
    {{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}},
    {{9},
     {4, 1, 5, 2, 4},
     {4, 1, 6, 3, 4},
     {4, 2, 6, 3, 5},
     {3, 1, 2, 6},
     {3, 2, 3, 4},
     {3, 1, 5, 3},
     {3, 1, 2, 3},
     {3, 4, 5, 6},
     {3, 4, 5, 6}}};
int L[333][2], R[333][2];
int Ans[333 * 333][10], cnt;
void addcir(int s, int l) {
  for (int i = (1); i <= (ans[s][0][0]); i++) {
    assert(ans[s][i][0]);
    ++cnt;
    for (int j = (0); j <= (ans[s][i][0]); j++)
      Ans[cnt][j] = ans[s][i][j] + (!!j) * l;
  }
}
void solve(int l, int r) {
  int s = r - l + 1;
  if (s <= 6) return addcir(s - 3, l);
  int x, y, c;
  if (s & 1) {
    x = 3, y = (s - 3) >> 1, c = 1;
    ++l;
    L[1][0] = l, L[1][1] = l + 1;
    L[2][0] = l, L[2][1] = l + 2;
    L[3][0] = l + 1, L[3][1] = l + 2;
    for (int i = (1); i <= (y); i++)
      for (int k = (0); k <= (1); k++) R[i][k] = l + 3 + 2 * (i - 1) + k;
    --l;
    addcir(0, l);
  } else {
    x = 2, y = (s - 4) >> 1, c = 2;
    ++l;
    for (int i = (1); i <= (x); i++)
      for (int k = (0); k <= (1); k++) L[i][k] = l + 2 * (i - 1) + k;
    for (int i = (1); i <= (y); i++)
      for (int k = (0); k <= (1); k++) R[i][k] = l + 4 + 2 * (i - 1) + k;
    --l;
    addcir(1, l);
  }
  for (int i = (1); i <= (x); i++)
    for (int j = (1); j <= (y); j++)
      for (int k = (1); k <= (c); k++)
        ++cnt, Ans[cnt][0] = 4, Ans[cnt][1] = L[i][0], Ans[cnt][2] = R[j][0],
               Ans[cnt][3] = L[i][1], Ans[cnt][4] = R[j][1];
  solve(l + ((s & 1) ? 3 : 4), r);
}
int main() {
  file();
  read(n);
  solve(0, n - 1);
  printf("%d\n", cnt);
  for (int i = (1); i <= (cnt); i++) {
    for (int j = (0); j <= (Ans[i][0]); j++) printf("%d ", Ans[i][j]);
    puts("");
  }
  return 0;
}