Problem: You are given a sequence a1, a2, ..., an consisting of different integers. It is required to split this sequence into the maximum number of subsequences such that after sorting integers in each of them in increasing order, the total sequence also will be sorted in increasing order.

Sorting integers in a subsequence is a process such that the numbers included in a subsequence are ordered in increasing order, and the numbers which are not included in a subsequence don't change their places.

Every element of the sequence must appear in exactly one subsequence.

Input

The first line of input data contains integer n (1 ≤ n ≤ 105) — the length of the sequence.

The second line of input data contains n different integers a1, a2, ..., an ( - 109 ≤ ai ≤ 109) — the elements of the sequence. It is guaranteed that all elements of the sequence are distinct.

Output

In the first line print the maximum number of subsequences k, which the original sequence can be split into while fulfilling the requirements.

In the next k lines print the description of subsequences in the following format: the number of elements in subsequence ci (0 < ci ≤ n), then ci integers l1, l2, ..., lci (1 ≤ lj ≤ n) — indices of these elements in the original sequence. 

Indices could be printed in any order. Every index from 1 to n must appear in output exactly once.

If there are several possible answers, print any of them.

Examples

Input

6
3 2 1 6 5 4


Output

4
2 1 3
1 2
2 4 6
1 5


Input

6
83 -75 -49 11 37 62


Output

1
6 1 2 3 4 5 6

Note

In the first sample output:

After sorting the first subsequence we will get sequence 1 2 3 6 5 4.

Sorting the second subsequence changes nothing.

After sorting the third subsequence we will get sequence 1 2 3 4 5 6.

Sorting the last subsequence changes nothing.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T, class U>
inline void relax_min(T &a, U b) {
  a = min(a, b);
}
template <class T, class U>
inline void relax_max(T &a, U b) {
  a = max(a, b);
}
template <typename T>
inline bool updmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T>
inline bool updmin(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
template <class T, class U>
istream &operator>>(istream &in, pair<T, U> &a) {
  in >> a.first >> a.second;
  return in;
}
template <class T, class U>
ostream &operator<<(ostream &out, const pair<T, U> &a) {
  out << a.first << ' ' << a.second;
  return out;
}
template <class T>
istream &operator>>(istream &in, vector<T> &a) {
  for (decltype(0)((i)) = (0); ((i)) < ((((int)(a).size()))); ++((i)))
    in >> a[i];
  return in;
}
template <class T>
ostream &operator<<(ostream &out, const vector<T> &a) {
  for (decltype(0)((i)) = (0); ((i)) < ((((int)(a).size()))); ++((i)))
    out << a[i] << ' ';
  return out;
}
int next_int() {
  int(tmp);
  scanf("%d", &(tmp));
  return tmp;
}
long long next_ll() {
  long long(tmp);
  scanf("%lld", &(tmp));
  return tmp;
}
long long modPow(long long a, long long b, long long MOD) {
  long long ret = 1;
  for (; b; b >>= 1) {
    if (b & 1)
      ret = ((((((long long)(ret) * (long long)(a))) % ((MOD))) + ((MOD))) %
             ((MOD)));
    a = ((((((long long)(a) * (long long)(a))) % ((MOD))) + ((MOD))) % ((MOD)));
  }
  return ret;
}
const long long EXACTLY_ONE_HUNDRED = 100;
const long long EXACTLY_ONE_THOUSAND = 1000;
const long long EXACTLY_ONE_MILLION = 1000 * 1000;
const long long EXACTLY_ONE_BILLION =
    EXACTLY_ONE_THOUSAND * EXACTLY_ONE_MILLION;
const int MAXN1 = (int)(EXACTLY_ONE_HUNDRED * EXACTLY_ONE_THOUSAND) + 3;
const int MAXN2 = (int)(2 * EXACTLY_ONE_HUNDRED * EXACTLY_ONE_THOUSAND) + 3;
const int MAXN3 = (int)(3 * EXACTLY_ONE_HUNDRED * EXACTLY_ONE_THOUSAND) + 3;
const int MAXN4 = (int)(4 * EXACTLY_ONE_HUNDRED * EXACTLY_ONE_THOUSAND) + 3;
const int MAXN5 = (int)(5 * EXACTLY_ONE_HUNDRED * EXACTLY_ONE_THOUSAND) + 3;
const int MAXN = MAXN3;
const int INF = (int)EXACTLY_ONE_BILLION + 7;
const int INF2 = 1 << 30;
const long long INFLL = 4LL * EXACTLY_ONE_BILLION * EXACTLY_ONE_BILLION + 7;
const long long INFLL2 = 1LL << 60;
const long double Pi = acos(-1.0);
const long double EPS = 1e-7;
const int MOD = (int)EXACTLY_ONE_BILLION + 7;
int main() {
  int(n);
  scanf("%d", &(n));
  vector<int> a(n);
  for (decltype(0)((IT_FOR_SCANNING)) = (0); ((IT_FOR_SCANNING)) < ((((n))));
       ++((IT_FOR_SCANNING)))
    scanf("%d", &(((a))[IT_FOR_SCANNING]));
  vector<pair<int, int> > b(n);
  for (decltype(0)((i)) = (0); ((i)) < ((n)); ++((i))) {
    b[i] = {a[i], i};
  }
  sort((b).begin(), (b).end());
  vector<vector<int> > ans;
  vector<int> used(n);
  for (decltype(0)((i)) = (0); ((i)) < ((n)); ++((i))) {
    vector<int> cur;
    for (int j = i; !used[j]; j = b[j].second) {
      cur.push_back(j + 1);
      used[j] = true;
    }
    if (((int)(cur).size())) {
      ans.push_back(cur);
    }
  }
  printf("%d\n", (((int)(ans).size())));
  for (decltype(0)((i)) = (0); ((i)) < ((((int)(ans).size()))); ++((i))) {
    printf("%d ", (((int)(ans[i]).size())));
    for (decltype(0)((IT_FOR_PRINTING)) = (0);
         ((IT_FOR_PRINTING)) < ((((int)(ans[i]).size())));
         ++((IT_FOR_PRINTING)))
      printf("%d ", ((ans[i])[IT_FOR_PRINTING]));
    printf("\n");
  }
  return 0;
}