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;
void printVP(vector<pair<long long int, long long int>> vec_of_pairs);
void printV(vector<int> vec);
void printVVI(vector<vector<int>> vec);
void printMapSP(map<string, pair<int, int>> mp);
void printPQ(priority_queue<int> pq);
void printVC(vector<char> vec);
void printCurrMap(map<char, set<char>> mp);
vector<int> curr_deck, prev_deck;
void updateDeck(int size, int curr_start, int prev_start) {
  for (int i = 0; i < size; i++) {
    curr_deck[curr_start + i] = prev_deck[prev_start + i];
  }
}
void solve() {
  int N, ci;
  cin >> N;
  for (int i = 0; i < N; i++) {
    cin >> ci;
    curr_deck.push_back(ci);
    prev_deck.push_back(ci);
  }
  vector<vector<int>> ans;
  bool is_sorted = false;
  while (!is_sorted) {
    int ind_1st_num, ind_2nd_num;
    bool found_consec_nums = false;
    for (int i = 0; i < N - 1; i++) {
      for (int j = i + 1; j < N; j++) {
        if (curr_deck[i] == curr_deck[j] + 1) {
          found_consec_nums = true;
          ind_1st_num = i;
          ind_2nd_num = j;
          break;
        }
      }
      if (found_consec_nums) break;
    }
    if (!found_consec_nums) {
      is_sorted = true;
    } else {
      int mid_ind = ind_1st_num;
      int num_check = curr_deck[ind_1st_num];
      while (curr_deck[mid_ind] == num_check) {
        num_check++;
        mid_ind++;
      }
      vector<int> curr_operation;
      curr_operation.push_back(0);
      int size_1 = ind_1st_num;
      if (size_1 != 0) {
        curr_operation[0]++;
        curr_operation.push_back(size_1);
        updateDeck(size_1, N - size_1, 0);
      }
      int size_2 = mid_ind - ind_1st_num;
      curr_operation[0]++;
      curr_operation.push_back(size_2);
      updateDeck(size_2, N - size_1 - size_2, size_1);
      int size_3 = ind_2nd_num - (mid_ind - 1);
      curr_operation[0]++;
      curr_operation.push_back(size_3);
      updateDeck(size_3, N - size_1 - size_2 - size_3, size_1 + size_2);
      int size_4 = (N - 1) - ind_2nd_num;
      if (size_4 != 0) {
        curr_operation[0]++;
        curr_operation.push_back(size_4);
        updateDeck(size_4, 0, size_1 + size_2 + size_3);
      }
      ans.push_back(curr_operation);
      prev_deck = curr_deck;
    }
  }
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i][0] << " ";
    for (int j = 1; j < ans[i].size(); j++) {
      cout << ans[i][j] << " ";
    }
    cout << endl;
  }
}
int main() {
  int t;
  t = 1;
  for (int i = 0; i < t; i++) {
    solve();
  }
  return 0;
}