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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
template <typename T>
inline void print(vector<T>& a) {
  for (int i = 0; i < a.size(); i++) cout << a[i] << " ";
  cout << "\n";
}
template <typename T>
inline void inp(vector<T>& a) {
  for (int i = 0; i < a.size(); i++) {
    cin >> a[i];
  }
}
template <typename T>
inline void out(T t) {
  cout << t << " ";
}
template <typename T, typename... A>
inline void out(T t, A... args) {
  cout << t << " ";
  out(args...);
}
int n;
vector<vector<int> > fans;
bool solve(vector<int>& c) {
  bool flag = 1;
  for (int i = 0; i < n; i++) {
    if (c[i] != i + 1) {
      flag = 0;
      break;
    }
  }
  if (flag) return 0;
  if (fans.size() >= n) return 0;
  vector<int> g(n, 0);
  for (int i = 1; i < n; i++) {
    if (c[i] == c[i - 1] + 1)
      g[i] = g[i - 1];
    else
      g[i] = g[i - 1] + 1;
  }
  int idx = 0;
  for (int i = 0; i < n; i++)
    if (c[i] != i + 1) {
      idx = i;
      break;
    }
  int x = c[idx];
  vector<int> ans;
  if (idx != 0) ans.push_back(idx);
  int gidx;
  for (int i = idx; i < n; i++) {
    if (c[i] == x - 1) {
      gidx = i;
      break;
    }
  }
  int group = g[gidx], tempx = 0, eidx;
  for (int i = idx; i < n; i++) {
    if (g[i] == group) {
      eidx = i;
      break;
    }
  }
  ans.push_back(eidx - idx);
  int pgroup = g[eidx], count = 1;
  for (int i = eidx + 1; i < n; i++) {
    if (g[i] == pgroup)
      count++;
    else {
      ans.push_back(count);
      count = 1;
      pgroup = g[i];
    }
  }
  if (count != 0) ans.push_back(count);
  fans.push_back(ans);
  vector<vector<int> > reversec;
  for (int i = 0, k = 0; i < ans.size() && k < n; i++) {
    vector<int> tempr;
    for (int j = 0; j < ans[i] && k < n; j++) {
      tempr.push_back(c[k++]);
    }
    reversec.push_back(tempr);
  }
  for (int i = reversec.size() - 1, k = 0; i >= 0 && k < n; i--) {
    for (int j = 0; j < reversec[i].size(); j++) {
      c[k++] = reversec[i][j];
    }
  }
  return 1;
}
int main() {
  cin >> n;
  vector<int> c(n);
  inp(c);
  while (solve(c)) {
    ;
  }
  cout << fans.size() << "\n";
  for (int i = 0; i < fans.size(); i++) {
    cout << fans[i].size() << " ";
    for (int j = 0; j < fans[i].size(); j++) {
      cout << fans[i][j] << " ";
    }
    cout << "\n";
  }
  return 0;
}