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;
constexpr int nax = 52;
int n, tab[nax + 1];
vector<vector<int> > ilo;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> tab[i];
  for (int i = 1; i < n; ++i) {
    int d = find(begin(tab) + 1, begin(tab) + n + 1, i) - begin(tab);
    ilo.push_back(vector<int>());
    if (i & 1) {
      if (d > 1) ilo.back().push_back(d - 1);
      if (n - d + 1 - (i - 1) > 0) ilo.back().push_back(n - d + 1 - (i - 1));
      for (int j = 1; j < i; ++j) ilo.back().push_back(1);
    } else {
      for (int j = 1; j < i; ++j) ilo.back().push_back(1);
      if (d - i + 1 > 0) ilo.back().push_back(d - i + 1);
      if (n - d > 0) ilo.back().push_back(n - d);
    }
    vector<vector<int> > arr;
    int sm = 1;
    for (int v : ilo.back()) {
      arr.push_back(vector<int>());
      for (int j = sm; j <= sm + v - 1; ++j) arr.back().push_back(tab[j]);
      sm += v;
    }
    reverse(begin(arr), end(arr));
    sm = 1;
    for (const auto& vv : arr)
      for (int v : vv) tab[sm++] = v;
  }
  if (n & 1) {
    ilo.push_back(vector<int>());
    for (int i = 1; i <= n; ++i) ilo.back().push_back(1);
    vector<vector<int> > arr;
    int sm = 1;
    for (int v : ilo.back()) {
      arr.push_back(vector<int>());
      for (int j = sm; j <= sm + v - 1; ++j) arr.back().push_back(tab[j]);
      sm += v;
    }
    reverse(begin(arr), end(arr));
    sm = 1;
    for (const auto& vv : arr)
      for (int v : vv) tab[sm++] = v;
  }
  int cnt = 0;
  for (const auto& vv : ilo)
    if (vv.size() > 1) ++cnt;
  cout << cnt << '\n';
  for (const auto& vv : ilo)
    if (vv.size() > 1) {
      cout << vv.size() << ' ';
      for (int v : vv) cout << v << ' ';
      cout << '\n';
    }
}