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;
template <typename T>
void setmax(T& x, T y) {
  x = max(x, y);
}
template <typename T>
void setmin(T& x, T y) {
  x = min(x, y);
}
template <typename T>
void print(const T& a, ostream& out) {
  for (auto i : a) out << i << ' ';
  out << endl;
}
const double PI = acos(-1.0);
const long long INF = 1e9 + 47;
const long long LINF = INF * INF;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
vector<int> go(vector<int> x) {
  vector<int> y;
  for (auto i : x)
    if (i) y.push_back(i);
  return y;
}
void roby(vector<int>& a, vector<int> x) {
  vector<vector<int>> b;
  vector<int> curr;
  int ptr = 0;
  for (auto i : a) {
    curr.push_back(i);
    if ((int)((curr).size()) == x[ptr]) {
      b.push_back(curr);
      curr.clear();
      ptr++;
    }
  }
  reverse(b.begin(), b.end());
  a.clear();
  for (auto i : b)
    for (auto j : i) a.push_back(j);
}
int is_shift(vector<int>& a, int n) {
  for (int i = (0); i < (n); ++i)
    if (a[i] == 0) {
      for (int j = (1); j < (n); ++j)
        if (a[(i + j) % n] != a[(i + j - 1) % n] + 1) {
          return -1;
        }
      return i;
    }
  assert(false);
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = (0); i < (n); ++i) cin >> a[i], a[i]--;
  vector<vector<int>> ans;
  while (true) {
    vector<int> pos(n);
    for (int i = (0); i < (n); ++i) pos[a[i]] = i;
    int id = -1;
    for (int i = (1); i < (n); ++i) {
      if (pos[i] < pos[i - 1]) {
        id = i;
        break;
      }
    }
    if (id == -1) {
      break;
    }
    int xxx = is_shift(a, n);
    if (xxx != -1) {
      ans.push_back({xxx, n - xxx});
      roby(a, ans.back());
      break;
    }
    int len = 1;
    while (a[pos[id] + len] == a[pos[id] + len - 1] + 1) len++;
    vector<int> x = go(
        {pos[id], len, pos[id - 1] - pos[id] - len + 1, n - pos[id - 1] - 1});
    ans.push_back(x);
    roby(a, ans.back());
  }
  assert((int)((ans).size()) <= n);
  cout << (int)((ans).size()) << endl;
  for (auto i : ans) {
    cout << (int)((i).size());
    for (auto j : i) cout << ' ' << j;
    cout << endl;
  }
  cerr << "Time elapsed: " << clock() / (double)CLOCKS_PER_SEC << endl;
  return 0;
}