Problem: You are given a deck of n cards numbered from 1 to n (not necessarily in this order in the deck). You have to sort the deck by repeating the following operation. 

  * Choose 2 ≤ k ≤ n and split the deck in k nonempty contiguous parts D_1, D_2,..., D_k (D_1 contains the first |D_1| cards of the deck, D_2 contains the following |D_2| cards and so on). Then reverse the order of the parts, transforming the deck into D_k, D_{k-1}, ..., D_2, D_1 (so, the first |D_k| cards of the new deck are D_k, the following |D_{k-1}| cards are D_{k-1} and so on). The internal order of each packet of cards D_i is unchanged by the operation. 



You have to obtain a sorted deck (i.e., a deck where the first card is 1, the second is 2 and so on) performing at most n operations. It can be proven that it is always possible to sort the deck performing at most n operations.

Examples of operation: The following are three examples of valid operations (on three decks with different sizes). 

  * If the deck is [3 6 2 1 4 5 7] (so 3 is the first card and 7 is the last card), we may apply the operation with k=4 and D_1=[3 6], D_2=[2 1 4], D_3=[5], D_4=[7]. Doing so, the deck becomes [7 5 2 1 4 3 6]. 
  * If the deck is [3 1 2], we may apply the operation with k=3 and D_1=[3], D_2=[1], D_3=[2]. Doing so, the deck becomes [2 1 3]. 
  * If the deck is [5 1 2 4 3 6], we may apply the operation with k=2 and D_1=[5 1], D_2=[2 4 3 6]. Doing so, the deck becomes [2 4 3 6 5 1]. 

Input

The first line of the input contains one integer n (1≤ n≤ 52) — the number of cards in the deck.

The second line contains n integers c_1, c_2, ..., c_n — the cards in the deck. The first card is c_1, the second is c_2 and so on.

It is guaranteed that for all i=1,...,n there is exactly one j∈\{1,...,n\} such that c_j = i.

Output

On the first line, print the number q of operations you perform (it must hold 0≤ q≤ n).

Then, print q lines, each describing one operation.

To describe an operation, print on a single line the number k of parts you are going to split the deck in, followed by the size of the k parts: |D_1|, |D_2|, ... , |D_k|. 

It must hold 2≤ k≤ n, and |D_i|≥ 1 for all i=1,...,k, and |D_1|+|D_2|+⋅⋅⋅ + |D_k| = n.

It can be proven that it is always possible to sort the deck performing at most n operations. If there are several ways to sort the deck you can output any of them.

Examples

Input


4
3 1 2 4


Output


2
3 1 2 1
2 1 3


Input


6
6 5 4 3 2 1


Output


1
6 1 1 1 1 1 1


Input


1
1


Output


0

Note

Explanation of the first testcase: Initially the deck is [3 1 2 4]. 

  * The first operation splits the deck as [(3) (1 2) (4)] and then transforms it into [4 1 2 3]. 
  * The second operation splits the deck as [(4) (1 2 3)] and then transforms it into [1 2 3 4]. 

Explanation of the second testcase: Initially the deck is [6 5 4 3 2 1]. 

  * The first (and only) operation splits the deck as [(6) (5) (4) (3) (2) (1)] and then transforms it into [1 2 3 4 5 6]. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
ostream& operator<<(ostream& out, vector<T>& v) {
  out << "[";
  for (auto k : v) out << k << " ";
  out << "]"
      << "\n";
  return out;
}
template <class T>
ostream& operator<<(ostream& out, set<T> s) {
  out << "{";
  for (auto k : s) out << k << " ";
  out << "}"
      << "\n";
  return out;
}
template <class T, class U>
ostream& operator<<(ostream& out, pair<T, U> p) {
  out << "[ " << p.first << " , " << p.second << " ] ";
  return out;
}
template <class T, class U>
istream& operator>>(istream& in, pair<T, U>& p) {
  in >> p.first >> p.second;
  return in;
}
vector<vector<int> > res;
int n;
void Ejecutar(vector<int> par, vector<int>& d) {
  vector<vector<int> > pars;
  vector<int> loc;
  int p = 0;
  for (auto k : par) {
    for (int j = 0; j < int(k); j++) {
      loc.push_back(d[p]);
      p++;
    }
    pars.push_back(loc);
    loc.resize(0);
  }
  reverse(pars.begin(), pars.end());
  vector<int> ld;
  for (auto k : pars)
    for (auto j : k) ld.push_back(j);
  d = ld;
}
void Paso(vector<int>& d, int i) {
  if (d[0] == 1) {
    int p = 0;
    while (d[p] != i) p++;
    if (p + 1 != i) {
      vector<int> par;
      for (int j = 0; j < int(i - 1); j++) par.push_back(1);
      par.push_back(p - i + 2);
      if (n - p - 1) par.push_back(n - p - 1);
      if (par.size() > 1) res.push_back(par);
      Ejecutar(par, d);
      return;
    }
  } else {
    int p = 0;
    while (d[n - p - 1] != i) p++;
    if (p + 1 != i) {
      vector<int> par;
      for (int j = 0; j < int(i - 1); j++) par.push_back(1);
      par.push_back(p - i + 2);
      if (n - p - 1) par.push_back(n - p - 1);
      reverse(par.begin(), par.end());
      if (par.size() > 1) res.push_back(par);
      Ejecutar(par, d);
      return;
    }
  }
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  cin >> n;
  vector<int> d(n);
  for (int i = 0; i < int(n); i++) cin >> d[i];
  for (int i = 0; i < int(n); i++) {
    if (d[i] == 1) {
      if (i != 0 && i != n - 1) {
        vector<int> p = {i, n - i};
        if (p.size() > 1) res.push_back(p);
        Ejecutar(p, d);
      }
      break;
    }
  }
  for (int i = int(2); i < int(n); i++) {
    Paso(d, i);
  }
  if (d[0] == n) {
    vector<int> r(n);
    for (int i = 0; i < int(n); i++) r[i] = 1;
    if (r.size() > 1) res.push_back(r);
    Ejecutar(r, d);
  }
  cout << res.size() << "\n";
  for (auto r : res) {
    cout << r.size() << " ";
    for (auto k : r) cout << k << " ";
    cout << "\n";
  }
  return 0;
}