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 change(vector<long long> &v, vector<long long> w) {
  vector<long long> k, t(v);
  vector<vector<long long> > p;
  while (!w.empty()) {
    long long u = w.back();
    w.pop_back();
    while (u--) {
      k.push_back(t.back());
      t.pop_back();
    }
    reverse(k.begin(), k.end());
    p.push_back(k);
    k.clear();
  }
  for (long long i = 0; i < (long long)p.size(); i++) {
    for (auto x : p[i]) k.push_back(x);
  }
  for (long long i = 0; i < (long long)v.size(); i++) {
    v[i] = k[i];
  }
}
void arrange(vector<long long> v, vector<long long> &s) {
  for (long long i = 0; i < (long long)v.size(); i++) {
    if (!i)
      s[i] = 0;
    else {
      s[i] = 0;
      if (v[i] == v[i - 1] + 1) s[i] += s[i - 1] + 1;
    }
  }
}
void setpos(vector<long long> v, vector<long long> &p) {
  for (long long i = 0; i < (long long)v.size(); i++) {
    p[v[i]] = i;
  }
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> v(n), p(n + 1), s(n, 0);
  for (long long i = 0; i < n; i++) cin >> v[i];
  setpos(v, p);
  arrange(v, s);
  vector<vector<long long> > ans;
  while (true) {
    vector<long long> w;
    for (long long i = 0; i < n; i++) {
      if (v[i] != n && p[v[i] + 1] < i) {
        if (p[v[i] + 1] == 0) {
          w.push_back(i - s[i] - p[v[i] + 1]);
        } else {
          w.push_back(p[v[i] + 1]);
          w.push_back(i - s[i] - p[v[i] + 1]);
        }
        w.push_back(s[i] + 1);
        if (i != n - 1) {
          w.push_back(n - 1 - i);
        }
      }
      if (!w.empty()) {
        change(v, w);
        setpos(v, p);
        arrange(v, s);
        ans.push_back(w);
        break;
      }
    }
    if (w.empty()) break;
    w.clear();
  }
  cout << (long long)ans.size() << '\n';
  ;
  for (auto x : ans) {
    cout << (long long)x.size() << " ";
    for (auto y : x) cout << y << " ";
    cout << '\n';
    ;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tc;
  tc = 1;
  for (int i = 1; i <= tc; i++) {
    solve();
  }
  return 0;
}