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;
const int MOD = 1e9 + 7;
const int MAX = 5e1 + 3;
int c[MAX];
int temp[MAX];
int N;
vector<vector<int> > ans;
void reverse(vector<int> v) {
  int start = 1;
  int end = N + 1;
  int size = v.size();
  for (int i = size - 1; i >= 0; i--) {
    int L = v[i];
    end -= L;
    for (int j = 0; j < L; j++) {
      temp[start + j] = c[end + j];
    }
    start += L;
  }
  for (int i = 1; i <= N; i++) {
    c[i] = temp[i];
  }
}
void solve() {
  cin >> N;
  for (int i = 1; i <= N; i++) {
    cin >> c[i];
  }
  int start = 0;
  int end = 0;
  int next = 0;
  for (int i = 1; i <= N; i++) {
    bool sorted = 1;
    bool reversed = 1;
    for (int j = 1; j <= N; j++) {
      if (c[j] != j) {
        sorted = 0;
      }
      if (c[N - j + 1] != j) {
        reversed = 0;
      }
    }
    if (sorted) {
      break;
    }
    if (reversed) {
      vector<int> v;
      for (int j = 1; j <= N; j++) {
        v.push_back(1);
      }
      ans.push_back(v);
      break;
    }
    for (int j = 1; j <= N; j++) {
      if (c[j] == i) {
        end = j;
      }
      if (c[j] == i + 1) {
        next = j;
      }
      if (c[j] == 1) {
        start = j;
      }
    }
    if (abs(end - next) == 1) {
      continue;
    }
    vector<int> v;
    if (next < end) {
      if (start < end) {
        v.push_back(next - 1);
        v.push_back(start - next);
        v.push_back(end - start + 1);
        v.push_back(N - end);
      } else {
        v.push_back(next - 1);
        v.push_back(end - next);
        for (int j = end; j <= start; j++) {
          v.push_back(1);
        }
        v.push_back(N - start);
      }
    } else {
      if (start < end) {
        v.push_back(start - 1);
        for (int j = start; j <= end; j++) {
          v.push_back(1);
        }
        v.push_back(next - end);
        v.push_back(N - next);
      } else {
        v.push_back(end - 1);
        v.push_back(start - end + 1);
        v.push_back(next - start);
        v.push_back(N - next);
      }
    }
    vector<int> v2;
    for (int a : v) {
      if (a > 0) {
        v2.push_back(a);
      }
    }
    reverse(v2);
    ans.push_back(v2);
  }
  cout << ans.size() << "\n";
  for (vector<int> v : ans) {
    cout << v.size();
    for (int a : v) {
      cout << " " << a;
    }
    cout << "\n";
  }
  return;
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}