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;
int n, p[69];
vector<vector<int>> ans;
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> p[i];
    --p[i];
  }
  vector<pair<int, int>> vec;
  for (int i = 0; i < n; ++i) {
    int j = i;
    while (j + 1 < n and p[j + 1] == p[j] + 1) ++j;
    vec.emplace_back(p[i], p[j]);
    i = j;
  }
  while (vec.size() > 1) {
    bool found = false;
    vector<pair<int, int>> new_vec, reversedVec;
    for (int i = 0; i < vec.size(); ++i) {
      for (int j = i + 1; j < vec.size(); ++j) {
        if (vec[i].first == vec[j].second + 1) {
          vector<int> new_move;
          int sz = 0;
          for (int k = 0; k < i; ++k) {
            sz += vec[k].second - vec[k].first + 1;
          }
          if (sz) new_move.emplace_back(sz);
          sz = 0;
          for (int k = i; k < j; ++k) {
            sz += vec[k].second - vec[k].first + 1;
          }
          if (sz) new_move.emplace_back(sz);
          sz = vec[j].second - vec[j].first + 1;
          if (sz) new_move.emplace_back(sz);
          sz = 0;
          for (int k = j + 1; k < vec.size(); ++k) {
            sz += vec[k].second - vec[k].first + 1;
          }
          if (sz) new_move.emplace_back(sz);
          ans.emplace_back(new_move);
          for (int k = j + 1; k < vec.size(); ++k) {
            reversedVec.emplace_back(vec[k]);
          }
          reversedVec.emplace_back(vec[j]);
          for (int k = i; k < j; ++k) {
            reversedVec.emplace_back(vec[k]);
          }
          for (int k = 0; k < i; ++k) {
            reversedVec.emplace_back(vec[k]);
          }
          found = true;
          break;
        }
      }
      if (found) break;
    }
    for (int i = 0; i < reversedVec.size(); ++i) {
      int j = i;
      pair<int, int> cur = reversedVec[i];
      while (j + 1 < reversedVec.size() and
             reversedVec[j + 1].first == cur.second + 1) {
        cur.second = reversedVec[j + 1].second;
        ++j;
      }
      new_vec.emplace_back(cur);
      i = j;
    }
    vec = new_vec;
  }
  cout << ans.size() << '\n';
  for (auto it : ans) {
    cout << it.size();
    for (int x : it) cout << " " << x;
    cout << '\n';
  }
  return 0;
}