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;
int n;
vector<int> a;
void print(vector<pair<int, int>> cur = {}) {
  cout << "a: ";
  for (int i : a) {
    cout << i << " ";
  }
  cout << endl;
  if (cur.size()) {
    cout << "cur: ";
    for (auto p : cur) {
      cout << "(" << p.first << ", " << p.second << ") ";
    }
    cout << endl;
  }
}
void change(vector<int> cur) {
  vector<vector<int>> b(cur.size());
  int p = 0;
  for (int i = 0; i < cur.size(); i++) {
    for (int j = p; j < cur[i] + p; j++) {
      b[i].push_back(a[j]);
    }
    p += cur[i];
  }
  reverse(b.begin(), b.end());
  vector<int> res;
  for (auto i : b) {
    for (auto j : i) {
      res.push_back(j);
    }
  }
  a = res;
}
vector<pair<int, int>> get() {
  int cnt = 1, b = a[0];
  vector<pair<int, int>> res;
  for (int i = 1; i < n; i++) {
    if (a[i] - a[i - 1] != 1) {
      res.push_back({b, cnt});
      cnt = 0;
      b = a[i];
    }
    cnt++;
  }
  res.push_back({b, cnt});
  vector<pair<int, int>> cur = res;
  sort(cur.begin(), cur.end());
  map<int, int> m;
  for (int i = 0; i < cur.size(); i++) {
    m[cur[i].first] = i + 1;
  }
  for (int i = 0; i < res.size(); i++) {
    res[i].first = m[res[i].first];
  }
  return res;
}
vector<vector<int>> ans;
void cl(vector<int>& cur) {
  vector<int> res;
  for (int i : cur) {
    if (i > 0) {
      res.push_back(i);
    }
  }
  cur = res;
}
void add(vector<int> cur) {
  cl(cur);
  change(cur);
  ans.push_back(cur);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int _ = 1;
  while (_--) {
    cin >> n;
    a.resize(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    vector<pair<int, int>> cur = get();
    while (cur.size() != 1) {
      int k = cur.size();
      vector<int> pos(k + 1);
      for (int i = 0; i < k; i++) {
        pos[cur[i].first] = i;
      }
      for (int i = 1; i < k; i++) {
        if (pos[i] < pos[i + 1]) {
          continue;
        }
        if (pos[i] == k - 1 && pos[i + 1] == 0) {
          add({cur[0].second, n - cur[0].second});
          break;
        }
        if (pos[i] == k - 1) {
          int s = 0;
          for (int j = 0; j < k; j++) {
            if (j < pos[i + 1]) {
              s += cur[j].second;
            }
          }
          add({s, n - s - cur[k - 1].second, cur[k - 1].second});
          break;
        }
        if (pos[i + 1] == 0) {
          int s = 0;
          for (int j = 0; j < k; j++) {
            if (j > pos[i]) {
              s += cur[j].second;
            }
          }
          add({cur[0].second, n - s - cur[0].second, s});
          break;
        }
        int s1 = 0, s2 = 0;
        for (int j = 0; j < k; j++) {
          if (j < pos[i + 1]) {
            s2 += cur[j].second;
          }
          if (j > pos[i]) {
            s1 += cur[j].second;
          }
        }
        add({s2, cur[pos[i + 1]].second, n - s2 - s1 - cur[pos[i + 1]].second,
             s1});
        break;
      }
      cur = get();
    }
    cout << ans.size() << endl;
    for (auto i : ans) {
      cout << i.size() << " ";
      for (int j : i) {
        cout << j << " ";
      }
      cout << endl;
    }
  }
}