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 <typename T>
void prarr(T a) {
  cerr << "[ ";
  for (auto i : a) {
    cerr << i << " ";
  }
  cerr << "]"
       << "\n";
}
template <typename Arg1>
void _er(const char* name, Arg1&& arg1) {
  cerr << "[" << name << " : " << arg1 << "]"
       << "\n";
}
template <typename Arg1, typename... Args>
void _er(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr << "[";
  cerr.write(names, comma - names) << " : " << arg1 << "] ";
  _er(comma + 2, args...);
}
vector<int> A, B;
vector<vector<int> > G1;
int n;
void doit() {
  for (int i = 2; i <= n; i++) {
    for (int j = i - 1; j < n; j++) {
      if (A[(j - i + 1)] == A[j] + 1) {
        vector<int> curr, curr1;
        int pos = (j - i + 1);
        while (pos + 1 < n && A[pos] + 1 == A[pos + 1]) {
          pos++;
        }
        int ma = max(0, j - i + 1);
        if (ma > 0) {
          curr1.push_back(ma);
        }
        ma = max(0, pos - (j - i + 1) + 1);
        if (ma > 0) {
          curr1.push_back(ma);
        }
        ma = max(0, j - (pos + 1) + 1);
        if (ma > 0) {
          curr1.push_back(ma);
        }
        ma = max(0, n - (j + 1));
        if (ma > 0) {
          curr1.push_back(ma);
        }
        G1.push_back(curr1);
        int k = j + 1;
        while (k < n) {
          curr.push_back(A[k]);
          k++;
        }
        k = pos + 1;
        while (k <= j && k < n) {
          curr.push_back(A[k]);
          k++;
        }
        k = (j - i + 1);
        while (k <= pos && k < n) {
          curr.push_back(A[k]);
          k++;
        }
        k = 0;
        while (k < (j - i + 1)) {
          curr.push_back(A[k]);
          k++;
        }
        A = curr;
        return;
      }
    }
  }
}
void solve() {
  A.clear();
  B.clear();
  G1.clear();
  cin >> n;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    A.push_back(a);
    B.push_back(a);
  }
  sort(B.begin(), B.end());
  while (true) {
    if (A == B) {
      break;
    }
    doit();
  }
  cout << G1.size() << "\n";
  for (auto it : G1) {
    vector<int> curr = it;
    cout << curr.size() << " ";
    for (auto jt : curr) {
      cout << jt << " ";
    }
    cout << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}