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;
ostream& operator<<(ostream& out, const vector<long long>& b) {
  for (auto k : b) out << k << " ";
  return out;
}
istream& operator>>(istream& in, pair<long long, long long>& b) {
  in >> b.first >> b.second;
  return in;
}
ostream& operator<<(ostream& out, const pair<long long, long long>& b) {
  out << "{" << b.first << ", " << b.second << "}";
  return out;
}
template <typename T1, typename T2>
inline bool chkmin(T1& x, const T2& y) {
  if (x > y) {
    x = y;
    return 1;
  }
  return 0;
}
template <typename T1, typename T2>
inline bool chkmax(T1& x, const T2& y) {
  if (x < y) {
    x = y;
    return 1;
  }
  return 0;
}
const long long INF = 1e18;
const long double PI = acos(-1);
mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());
bool good(const vector<long long>& a) {
  for (long long i = 1; i < a.size(); i++) {
    if (a[i] - a[i - 1] != 1) {
      return 0;
    }
  }
  return 1;
}
void make(vector<long long>& a, vector<long long>& blocks) {
  long long ind = 0;
  long long sz = blocks.size();
  vector<vector<long long>> w(sz);
  for (long long i = 0; i < sz; i++) {
    for (long long j = 0; j < blocks[i]; j++) {
      w[i].push_back(a[ind]);
      ind++;
    }
  }
  a.clear();
  for (long long i = sz - 1; i >= 0; i--) {
    for (auto k : w[i]) {
      a.push_back(k);
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<vector<long long>> ans;
  while (!good(a)) {
    vector<long long> pos(n + 1, -1);
    vector<long long> cur;
    bool ok = 0;
    for (long long i = n - 1; i >= 0; i--) {
      if (pos[a[i] - 1] != -1 && !ok) {
        long long ind = pos[a[i] - 1];
        while (ind > 0 && a[ind - 1] == a[ind] - 1) {
          ind--;
        }
        if (i > 0) {
          cur.push_back(i);
        }
        cur.push_back(ind - i);
        cur.push_back(pos[a[i] - 1] - ind + 1);
        if (pos[a[i] - 1] < n - 1) {
          cur.push_back(n - pos[a[i] - 1] - 1);
        }
        ans.push_back(cur);
        make(a, cur);
        ok = 1;
      }
      pos[a[i]] = i;
    }
  }
  cout << ans.size() << "\n";
  for (auto k : ans) {
    cout << k.size() << " " << k << "\n";
  }
}