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>
#pragma GCC optimize("Ofast")
using namespace std;
const long long MAX = 1e18 + 1;
const double pi = 3.1415926535897932384626433832;
const long long M = 1e9 + 7;
long long modpow(long long x, long long n, long long mod1) {
  if (n == 0) return 1 % mod1;
  long long u = modpow(x, n / 2, mod1);
  u = (u * u) % mod1;
  if (n % 2 == 1) u = (u * x) % mod1;
  return u;
}
long long POW(long long x, long long y) { return modpow(x, y, LONG_LONG_MAX); }
long long MI(long long x, long long mod = M) { return modpow(x, mod - 2, mod); }
long long gcd(long long x, long long y) {
  if (x == 0)
    return y;
  else
    return gcd(y % x, x);
}
long long lcm(long long x, long long y) { return (x * y) / gcd(x, y); }
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6) {
    if (n % i == 0 || n % (i + 2) == 0) {
      return false;
    }
  }
  return true;
}
bool sumdigits(long long a) {
  long long result = 0;
  while (a > 0) {
    result += a % 10;
    a /= 10;
  }
  return result;
}
const long long Node = 2e5 + 7;
vector<long long> adj[Node];
int vis[Node];
void addedge(long long u, long long v) {
  adj[u - 1].push_back(v - 1);
  adj[v - 1].push_back(u - 1);
}
long long ans = 0;
void dfs(long long u, vector<long long>& v,
         vector<pair<long long, long long>>& v1) {
  vis[u] = 1;
  v.push_back(u);
  if (!vis[v1[u].second])
    dfs(v1[u].second, v, v1);
  else {
    ans++;
    return;
  }
}
long long modInverse(long long a, long long m) {
  long long m0 = m;
  long long y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long q = a / m;
    long long t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
const long long N = 2e5 + 7;
vector<long long> v[N];
void solve() {
  long long n;
  cin >> n;
  vector<pair<long long, long long>> v1;
  long long a[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    v1.push_back({a[i], i});
  }
  sort(v1.begin(), v1.end());
  for (long long i = 0; i < n; i++) {
    if (!vis[i]) {
      dfs(i, v[ans], v1);
    }
  }
  cout << ans << endl;
  for (long long i = 0; i < ans; i++) {
    cout << v[i].size() << " ";
    for (auto x : v[i]) cout << x + 1 << " ";
    cout << endl;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}