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;
vector<vector<int>> x;
void change2(int n, vector<int> &a, vector<int> k) {
  deque<int> y, z;
  for (int i = 0; i < n; i++) {
    y.push_back(a[i]);
  }
  for (int i : k) {
    deque<int> t;
    for (int j = 0; j < i; j++) {
      t.push_back(y.front());
      y.pop_front();
    }
    while (t.size()) {
      z.push_front(t.back());
      t.pop_back();
    }
  }
  for (int i = 0; i < n; i++) {
    a[i] = z[i];
  }
}
void change(int i, int j, int n, vector<int> &a) {
  int e = i + 1;
  while (e < n && a[e] - a[e - 1] == 1) {
    e++;
  }
  e--;
  vector<int> k;
  if (i > 0) {
    k.push_back(i);
  }
  k.push_back(e - i + 1);
  k.push_back(j - e);
  if (j + 1 < n) {
    k.push_back(n - 1 - j);
  }
  x.push_back(k);
  change2(n, a, k);
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  while (true) {
    bool f = false;
    for (int i = 0; i < n && !f; i++) {
      for (int j = i + 1; j < n && !f; j++) {
        if (a[i] - a[j] == 1) {
          f = true;
          change(i, j, n, a);
        }
      }
    }
    if (!f) {
      break;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  solve();
  cout << (int)(x.size()) << endl;
  for (int i = 0; i < x.size(); i++) {
    cout << (int)(x[i].size()) << ' ';
    for (int j : x[i]) {
      cout << j << ' ';
    }
    cout << endl;
  }
  return 0;
}