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 maxn = 1e6 + 100;
vector<vector<int>> g;
vector<int> c;
vector<int> find(vector<int> v) {
  vector<int> ans;
  for (auto& t : v)
    if (t) ans.push_back(t);
  return ans;
}
void go(vector<int> h) {
  h = find(h);
  g.push_back(h);
  vector<vector<int>> parts;
  int cnt = 0;
  for (int i = 0; i < h.size(); i++) {
    parts.emplace_back();
    for (int j = 0; j < h[i]; j++) parts.back().push_back(c[cnt++]);
  }
  reverse(parts.begin(), parts.end());
  vector<int> res;
  for (auto& x : parts)
    for (auto& xx : x) res.push_back(xx);
  c = res;
}
int main() {
  int t = 1;
  while (t--) {
    int n;
    cin >> n;
    c.resize(n);
    for (int i = 0; i < n; i++) cin >> c[i];
    while (1) {
      bool f = 0;
      for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++)
          if (c[j] + 1 == c[i]) {
            if (f) break;
            int l = i, m = i, r = j;
            while (c[m + 1] == c[m] + 1) m++;
            vector<int> s;
            s.push_back(l);
            s.push_back(m - l + 1);
            s.push_back(r - m);
            s.push_back(n - r - 1);
            f = 1;
            go(s);
          }
      if (!f) break;
    }
    cout << g.size() << "\n";
    for (auto& x : g) {
      cout << x.size();
      for (auto& xx : x) cout << " " << xx;
      cout << "\n";
    }
  }
  return 0;
}