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;
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
const long long maxn = 100;
long long mas[maxn], mas2[maxn];
vector<pair<long long, long long> > cur;
vector<pair<long long, long long> > ans[maxn];
void split(long long n) {
  cur.clear();
  long long a, b;
  a = 1, b = 1;
  for (long long i = 2; i <= n; i++) {
    if (mas[i] == mas[i - 1] + 1) {
      b = i;
    } else {
      cur.push_back({a, b});
      a = i, b = i;
    }
  }
  cur.push_back({a, b});
}
signed main() {
  fast();
  long long n, i, j, p, k, a, b;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> mas[i];
  }
  long long x;
  p = 0;
  while (true) {
    split(n);
    long long sz = cur.size();
    if (sz == 1) {
      break;
    }
    p++;
    x = mas[n];
    if (x == n) {
      x = mas[cur[sz - 2].second];
      for (i = 1; i <= n; i++) {
        if (mas[i] == x + 1) {
          if (i != 1) {
            ans[p].push_back({1, i - 1});
          }
          ans[p].push_back({i, cur[sz - 2].first - 1});
          ans[p].push_back(cur[sz - 2]);
          ans[p].push_back(cur[sz - 1]);
          k = 1;
          for (j = cur[sz - 1].first; j <= cur[sz - 1].second; j++) {
            mas2[k] = mas[j];
            k++;
          }
          for (j = cur[sz - 2].first; j <= cur[sz - 2].second; j++) {
            mas2[k] = mas[j];
            k++;
          }
          for (j = i; j <= cur[sz - 2].first - 1; j++) {
            mas2[k] = mas[j];
            k++;
          }
          if (i != 1) {
            for (j = 1; j <= i - 1; j++) {
              mas2[k] = mas[j];
              k++;
            }
          }
          for (k = 1; k <= n; k++) {
            mas[k] = mas2[k];
          }
          break;
        }
      }
    } else {
      for (i = 1; i <= n; i++) {
        if (mas[i] == x + 1) {
          if (i != 1) {
            ans[p].push_back({1, i - 1});
          }
          ans[p].push_back({i, cur[sz - 1].first - 1});
          ans[p].push_back(cur[sz - 1]);
          k = 1;
          for (j = cur[sz - 1].first; j <= cur[sz - 1].second; j++) {
            mas2[k] = mas[j];
            k++;
          }
          for (j = i; j <= cur[sz - 1].first - 1; j++) {
            mas2[k] = mas[j];
            k++;
          }
          if (i != 1) {
            for (j = 1; j <= i - 1; j++) {
              mas2[k] = mas[j];
              k++;
            }
          }
          for (k = 1; k <= n; k++) {
            mas[k] = mas2[k];
          }
          break;
        }
      }
    }
  }
  cout << p << endl;
  for (i = 1; i <= p; i++) {
    cout << ans[i].size() << " ";
    for (j = 0; j < ans[i].size(); j++) {
      cout << ans[i][j].second - ans[i][j].first + 1 << " ";
    }
    cout << endl;
  }
  return 0;
}