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 fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
void ran() { srand(chrono::steady_clock::now().time_since_epoch().count()); }
long long get_rand() {
  long long a = rand();
  long long b = rand();
  return a * (RAND_MAX + 1ll) + b;
}
void usaco() {
  freopen("problem.in", "r", stdin);
  freopen("problem.out", "w", stdout);
}
vector<vector<int>> ops;
int arr[55];
int narr[55];
int n;
void enact(vector<int> li) {
  ops.push_back(li);
  int L = 1;
  vector<vector<int>> tr;
  for (auto x : li) {
    int R = L + x - 1;
    vector<int> l;
    for (int j = L; j <= R; j++) l.push_back(j);
    tr.push_back(l);
    L = R + 1;
  }
  reverse(tr.begin(), tr.end());
  int cr = 1;
  for (auto m : tr) {
    for (auto l : m) {
      narr[cr] = arr[l];
      cr++;
    }
  }
  for (int i = 1; i <= n; i++) arr[i] = narr[i];
}
void merger(vector<pair<int, pair<int, int>>> blocks, int l, int r) {
  vector<int> ls;
  int ad = 0;
  for (int i = 0; i < blocks.size(); i++) {
    if (i <= l)
      ls.push_back(blocks[i].first);
    else if (i > r)
      ls.push_back(blocks[i].first);
    else {
      ad += blocks[i].first;
      if (i == r) ls.push_back(ad);
    }
  }
  enact(ls);
}
signed main() {
  ran();
  fast();
  cin >> n;
  if (n == 1) {
    cout << 0 << endl;
    exit(0);
  }
  for (int i = 1; i <= n; i++) cin >> arr[i];
  bool notk = true;
  while (notk) {
    vector<pair<int, pair<int, int>>> blocks;
    blocks.push_back({1, {arr[1], arr[1]}});
    for (int i = 2; i <= n; i++) {
      if (arr[i] == blocks.back().second.second + 1) {
        blocks.back().second.second = arr[i];
        blocks.back().first++;
      } else {
        blocks.push_back({1, {arr[i], arr[i]}});
      }
    }
    bool need = true;
    for (int i = 0; i < blocks.size(); i++) {
      for (int j = i + 1; j < blocks.size(); j++) {
        if (need && blocks[i].second.first == blocks[j].second.second + 1) {
          need = false;
          merger(blocks, i, j);
        }
      }
    }
    notk = false;
    for (int i = 1; i <= n; i++) {
      if (arr[i] != i) notk = true;
    }
  }
  cout << ops.size() << endl;
  for (auto x : ops) {
    cout << x.size() << ' ';
    for (auto j : x) cout << j << ' ';
    cout << endl;
  }
}