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;
using i8 = char;
using ui8 = unsigned char;
using i16 = short;
using u16 = unsigned short;
using i32 = int;
using ui32 = unsigned int;
using i64 = long long int;
using ui64 = unsigned long long int;
const i64 INF = 1e18;
vector<int> input;
const int N = 100100;
struct TTask {
  int n;
  vector<int> a;
  vector<string> ans;
  void Solve() {
    cin >> n;
    a.resize(n);
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
    }
    while (!is_sorted(a.begin(), a.end())) {
      vector<pair<int, int>> groups;
      set<int> all;
      int first = -1;
      int second = -1;
      for (int i = 0; i < n; ++i) {
        if (all.count(a[i] + 1)) {
          first = i;
          second = find(a.begin(), a.end(), a[i] + 1) - a.begin();
        }
        all.insert(a[i]);
      }
      if (first == -1) {
        (void(1));
        ;
        assert(false);
      }
      int start = first;
      while (a[start] - 1 == a[start - 1]) {
        --start;
      }
      vector<int> vv;
      if (second != 0) {
        vv.push_back(second);
      }
      vv.push_back(start - second);
      vv.push_back(first - start + 1);
      if (first != n - 1) {
        vv.push_back(n - first - 1);
      }
      stringstream ss;
      ss << vv.size() << ' ';
      for (int x : vv) {
        ss << x << ' ';
      }
      ans.push_back(ss.str());
      vector<int> b;
      for (int i = first + 1; i < n; ++i) {
        b.push_back(a[i]);
      }
      for (int i = start; i <= first; ++i) {
        b.push_back(a[i]);
      }
      for (int i = second; i < start; ++i) {
        b.push_back(a[i]);
      }
      for (int i = 0; i < second; ++i) {
        b.push_back(a[i]);
      }
      a = b;
    }
    cout << ans.size() << endl;
    for (auto& s : ans) {
      cout << s << endl;
    }
  }
};
int main() {
  srand(time(0));
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout.precision(15);
  cout.setf(ios::fixed);
  cerr.precision(15);
  cerr.setf(ios::fixed);
  int t = 1;
  for (int i = 0; i < t; ++i) {
    TTask task = {};
    task.Solve();
  }
  return 0;
}